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CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Differential Equations
Quiz 8
The solution of
y
e
−
x
/
y
d
x
−
(
x
e
(
−
x
/
y
)
+
y
3
)
d
y
=
0
is
Report Question
0%
e
−
x
/
y
+
y
2
=
C
0%
x
e
−
x
/
y
+
y
=
C
0%
2
e
−
x
/
y
+
y
2
=
C
0%
e
−
x
/
y
+
2
y
2
=
C
Explanation
y
e
−
x
/
y
d
x
−
(
x
e
−
x
/
y
+
y
3
)
d
y
=
0
⟹
(
y
d
x
−
x
d
y
)
e
−
x
/
y
−
y
3
d
y
=
0
⟹
y
d
x
−
x
d
y
y
2
e
−
x
/
y
=
y
d
y
⟹
d
(
x
/
y
)
e
−
x
/
y
=
y
d
y
−
e
−
x
/
y
=
y
2
2
+
C
2
e
−
x
/
y
+
y
2
=
C
The solution of the following differential equation
[
1
+
x
√
(
x
2
+
y
2
)
]
d
x
+
[
√
(
x
2
+
y
2
)
−
1
]
y
d
y
=
0
is equal to
Report Question
0%
x
2
+
y
2
2
+
1
3
(
x
2
+
y
2
)
3
/
2
=
c
0%
x
−
y
3
2
+
1
3
(
x
2
+
y
2
)
3
/
2
=
c
0%
x
−
y
2
2
+
1
3
(
x
2
+
y
2
)
3
/
2
=
c
0%
None of these
Explanation
[
1
+
x
√
(
x
2
+
y
2
)
]
d
x
+
[
√
(
x
2
+
y
2
)
−
1
]
y
d
y
=
0
⟹
d
x
−
y
d
y
+
√
(
x
2
+
y
2
)
(
x
d
x
+
y
d
y
)
=
0
⟹
d
x
−
y
d
y
+
1
2
√
(
x
2
+
y
2
)
d
(
x
2
+
y
2
)
=
0
Integrating, we have
x
−
y
2
2
+
1
2
∫
√
t
d
t
=
c
,
(
t
=
√
(
x
2
+
y
2
)
)
or
x
−
y
2
2
+
1
3
(
x
2
+
y
2
)
3
/
2
)
=
c
The solution of the differential equation
d
y
d
x
=
3
x
2
y
4
+
2
x
y
x
2
−
2
x
3
y
3
is
Report Question
0%
y
2
x
−
x
3
y
2
=
c
0%
x
2
y
2
+
x
3
y
3
=
c
0%
x
2
y
+
x
3
y
2
=
c
0%
x
2
3
y
−
2
x
3
y
2
=
c
Explanation
Re-write the D.E. as
(
2
x
y
d
x
−
x
2
d
y
)
+
y
2
(
3
x
2
y
2
d
x
+
2
x
3
y
d
y
)
=
0
Dividing by
y
2
, we get
y
2
x
d
x
−
x
2
d
y
y
2
+
y
2
3
x
2
d
x
+
x
3
2
y
d
y
=
0
or
d
(
x
2
y
)
+
d
(
x
3
y
2
)
=
0
Integrating, we get the solution
x
2
y
+
x
3
y
2
=
c
The solution of the differential equation
x
2
d
y
d
x
cos
1
x
−
y
sin
1
x
=
−
1
,
where
y
→
−
1
as
x
→
∞
is
Report Question
0%
y
=
sin
1
x
−
cos
1
x
0%
y
=
x
+
1
x
sin
1
x
0%
y
=
cos
1
x
+
sin
1
x
0%
y
=
x
+
1
x
cos
1
/
x
Explanation
x
2
d
y
d
x
cos
1
x
−
y
sin
1
x
=
−
1
⟹
d
y
d
x
−
y
x
2
tan
1
x
=
−
sec
1
x
1
x
2
(linear)
I.F.
=
e
∫
1
x
2
tan
1
x
d
x
=
sec
1
x
⟹
solutions is
y
sec
1
x
=
−
∫
sec
2
(
1
x
)
1
x
2
d
x
=
tan
1
x
+
c
Given
y
→
−
1
,
x
→
∞
⟹
x
=
−
1
Hence equation of curve is
y
=
sin
1
x
−
cos
1
x
The solution of the differential equation
y
(
2
x
2
+
y
)
d
y
d
x
=
(
1
−
4
x
y
2
)
x
2
is given by
Report Question
0%
3
(
x
2
y
)
2
+
y
3
−
x
3
=
c
0%
x
y
2
+
y
3
3
−
x
3
3
=
0
0%
2
5
y
x
5
+
y
3
3
=
x
3
3
−
4
x
y
3
3
+
c
0%
None of these
Explanation
y
(
2
x
4
+
y
)
d
y
d
x
=
91
−
4
x
y
2
)
x
2
⟹
2
x
4
y
d
y
+
y
2
d
y
+
4
x
3
y
2
d
x
−
x
2
d
x
=
0
⟹
2
x
2
y
(
x
2
d
y
+
2
x
y
d
x
)
+
y
2
d
y
−
x
2
d
x
=
0
⟹
2
x
2
y
d
(
x
2
y
)
+
y
2
d
y
−
x
2
d
x
=
0
Integrating, we get
(
x
2
y
2
)
2
+
y
3
3
−
x
3
3
=
c
or
3
(
x
2
y
)
2
+
y
3
−
x
3
=
c
Solution of the differential equation
(
y
+
x
√
x
y
(
x
+
y
)
)
d
x
+
(
y
√
x
y
(
x
+
y
)
−
x
)
d
y
=
0
is
Report Question
0%
x
2
+
y
2
2
+
tan
−
1
√
y
x
=
c
0%
x
2
+
y
2
2
+
tan
−
1
√
x
y
=
c
0%
x
2
+
y
2
2
+
2
cot
−
1
√
x
y
=
c
0%
None of these
Explanation
The given equation is written as
y
d
x
−
x
d
y
+
x
√
x
y
(
x
+
y
)
d
x
+
y
√
x
y
(
x
+
y
)
d
y
=
0
⟹
y
d
x
−
x
d
y
+
(
x
+
y
)
√
x
y
(
x
d
x
+
y
d
y
)
=
0
⟹
y
d
x
−
x
d
y
y
2
+
(
x
y
+
1
)
√
x
y
(
d
(
x
2
+
y
2
2
)
)
=
0
⟹
d
x
2
+
y
2
2
+
d
(
x
y
)
(
x
y
+
1
)
√
x
y
=
0
⟹
x
2
+
y
2
2
+
2
t
a
n
−
1
√
x
y
=
c
Number of values of
m
∈
N
for which
y
=
e
m
x
is a solution of the differential equation
d
3
y
d
x
3
−
3
d
2
y
d
x
2
−
4
d
y
d
x
+
12
y
=
0
Report Question
0%
0
0%
1
0%
2
0%
More than
2
Explanation
y
=
e
m
x
satisfies
d
3
y
d
x
3
−
3
d
2
y
d
x
2
−
4
d
y
d
x
+
12
y
=
0
then
e
m
x
(
m
3
−
3
m
2
−
4
m
+
12
)
=
0
⟹
m
=
±
2
,
3
m
∈
N
hence
m
∈
(
2
,
3
)
The solution of,
y
d
x
−
x
d
y
+
(
1
+
x
2
)
d
x
+
x
2
sin
y
d
y
=
0
,
Report Question
0%
x
+
1
−
y
2
+
cos
y
+
C
=
0
0%
y
+
1
−
x
2
+
cos
y
+
C
=
0
0%
x
y
+
1
y
−
y
+
cos
y
+
C
=
0
0%
y
x
+
1
x
−
x
+
cos
y
+
C
=
0
Report Question
0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect
Explanation
y
=
c
1
cos
2
x
+
c
2
sin
2
x
+
c
3
cos
2
x
+
c
4
e
2
x
+
c
5
e
2
x
y
=
c
1
cos
2
x
+
c
2
[
1
−
cos
2
x
2
]
+
c
3
[
cos
2
x
+
1
2
]
+
c
4
e
2
x
+
c
5
e
2
x
=
(
c
1
−
c
2
2
+
c
3
2
)
cos
2
x
+
(
c
2
2
+
c
3
2
)
+
(
c
4
+
c
5
)
e
2
x
=
λ
1
cos
2
x
+
λ
2
e
2
x
+
λ
3
⇒
Total number of independent parameters in the given general solution is
3
.
The solution of
d
y
d
x
+
x
sin
2
y
=
x
3
cos
2
y
, is
Report Question
0%
e
x
2
=
(
x
2
−
1
)
e
x
2
tan
y
+
C
0%
e
x
2
tan
y
=
1
2
(
x
2
−
1
)
e
x
2
+
C
0%
e
x
2
tan
y
=
(
x
2
−
1
)
tan
y
+
C
0%
N
o
n
e
o
f
t
h
e
s
e
Differential equation
d
y
d
x
=
f
(
x
)
g
(
x
)
can be solved by
separating variable
d
y
g
(
y
)
=
f
(
x
)
d
x
The equation of the curve to the point
(
1
,
0
)
which
stratifies the differential equation
(
1
+
y
2
)
d
x
−
x
y
d
y
=
0
Report Question
0%
x
2
+
y
2
=
1
0%
x
2
−
y
2
=
1
0%
x
2
+
y
2
=
2
0%
x
2
−
y
2
=
2
Explanation
d
x
x
=
y
d
y
1
+
y
2
⇒
ln
x
=
1
2
.
ln
(
1
+
y
2
)
+
C
2
⇒
2
ln
x
=
ln
(
1
+
y
2
)
+
C
⇒
ln
x
2
=
ln
(
1
+
y
2
)
+
C
From the given condition,
(
x
,
y
)
=
(
1
,
0
)
,
C
=
0
⇒
ln
x
2
=
ln
(
1
+
y
2
)
⇒
x
2
=
1
+
y
2
∴
x
2
−
y
2
=
1
The solution of the differential equation
(
e
x
2
+
e
y
2
)
y
d
y
d
x
+
e
x
2
(
x
y
2
−
x
)
=
0
is
Report Question
0%
e
x
2
(
y
2
−
1
)
+
e
y
2
=
C
0%
e
y
2
(
x
2
−
1
)
+
e
x
2
=
C
0%
e
y
2
(
y
2
−
1
)
+
e
x
2
=
C
0%
e
x
2
(
y
−
1
)
+
e
y
2
=
C
Explanation
y
2
=
t
;
2
y
d
y
d
x
;
hence the differential equation becomes
(
e
x
2
+
e
t
)
+
2
e
x
62
(
x
t
−
x
)
=
0
e
x
2
+
e
t
+
2
e
x
2
x
(
t
−
1
)
d
x
d
t
=
0
put
e
x
2
=
z
;
e
x
2
2
x
d
x
d
t
=
d
z
d
t
⟹
d
z
d
t
+
z
9
t
−
1
)
=
e
t
(
t
−
1
)
;
I.F.
=
e
∫
d
t
t
−
1
=
e
ln
(
t
−
1
)
=
t
−
1
⟹
z
(
t
−
1
)
=
−
∫
(
e
t
)
d
t
⟹
z
(
t
−
1
)
=
e
t
+
C
⟹
e
x
2
(
y
2
−
1
)
=
−
e
y
2
+
C
⟹
e
x
2
(
y
2
−
1
)
+
e
y
2
=
C
y
satisfies the differential equation
Report Question
0%
d
y
d
x
+
y
=
e
x
(
cos
x
−
sin
x
)
−
e
−
x
(
cos
x
−
sin
x
)
0%
d
y
d
x
−
y
=
e
x
(
cos
x
−
sin
x
)
+
e
−
x
(
cos
x
+
sin
x
)
0%
d
y
d
x
+
y
=
e
x
(
cos
x
+
sin
x
)
−
e
−
x
(
cos
x
−
sin
x
)
0%
d
y
d
x
−
y
=
e
x
(
cos
x
−
sin
x
)
+
e
−
x
(
cos
x
−
sin
x
)
Explanation
f
(
0
)
=
2
f
(
x
)
=
(
e
x
+
e
−
x
)
cos
x
−
2
x
−
[
x
∫
x
0
f
′
(
t
)
d
t
−
∫
0
x
t
1
f
′
(
t
)
0
d
t
]
f
(
x
)
=
(
e
x
+
e
−
x
)
cos
x
−
2
x
−
[
x
(
f
(
x
)
−
f
(
0
)
)
−
t
.
f
(
t
)
x
0
−
∫
x
0
f
(
t
)
d
t
]
f
(
x
)
=
(
e
x
+
e
−
x
)
cos
x
−
2
x
−
x
f
(
x
)
+
2
x
+
[
x
f
(
x
)
−
∫
x
0
f
(
t
)
d
t
]
f
(
x
)
=
(
e
x
+
e
−
x
)
cos
x
−
∫
x
0
f
(
t
)
d
t
...
(
i
)
On differentiating Eq.
(
i
)
f
′
(
x
)
+
f
(
x
)
=
cos
x
(
e
x
−
e
−
x
)
−
(
e
x
−
e
−
x
)
sin
x
...
(
i
i
)
Hence,
d
y
d
x
+
y
=
e
x
(
cos
x
−
sin
x
)
−
e
−
x
(
cos
x
+
sin
x
)
Solution of the diffrential equation
d
y
d
x
+
1
+
y
2
√
1
−
x
2
=
0
is
Report Question
0%
tan
−
1
y
+
sin
−
1
x
=
C
0%
tan
−
1
x
+
sin
−
1
y
=
C
0%
tan
−
1
x
.
sin
−
1
y
=
C
0%
tan
−
1
x
−
sin
−
1
y
=
C
Explanation
d
y
d
x
+
1
+
y
2
√
1
−
x
2
=
0
\displaystyle \Rightarrow \dfrac{dy}{1+y^2}+ \dfrac {dx}{\sqrt {1-x^2}}=0
\Rightarrow \tan^{-1} y + \sin ^{-1} x=C
Differential equation
\dfrac{dy}{dx}=f(x)g(x)
can be solved by
separating variable
\dfrac{dy}{g(y)}=f(x)dx
If
\dfrac{dy}{dx}=1+x+y+xy
and
y(-1)=0
, then
y
is equal to
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e^{\dfrac{(1-x)^1}{2}}
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e^{\dfrac{(1+x)^2}{2}} -1
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In
(1+x)-1
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1+x
Explanation
\dfrac{dy}{dx}=1+x+y+xy
\dfrac{dy}{dx}=(1+x).(1+y)
\dfrac {dy}{(1+y)}=(1+x)dx
\ln (1+y)=x+\dfrac {x^2}{2}+c
at
x=-1, y=0
\ln (1+0)=-1+\dfrac {1}{2}+c
0=\dfrac {-1}{2}+c \Rightarrow c=\dfrac12
\ln (1+y)=x+\dfrac {x^2}{2}+\dfrac 12
\ln (1+y)=\dfrac {(x+1)^2}{2}
y=e^{\dfrac {(x+1)^2}{2}}-1
The solution of the differential equation
\dfrac{x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+ \dots}{1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dots}=\dfrac{dx-dy}{dx+dy}
is
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2ye^{2x}=Ce^{2x}+1
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2ye^{2x}=Ce^{2x}-1
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ye^{2x}=Ce^{2x}+2
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None of these
Explanation
Applying componendo and divedendo we get
\dfrac{dy}{dx}=\dfrac{e^{-x}}{e^x}=e^{-2x}
\implies 2y=-e^{-2x}+C
\implies 2ye^{2x}=Ce^{2x}-1
Integrating factor of the differential equation
(1-x^2)\dfrac {dy}{dx}-xy=1
is
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-x
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\dfrac {x}{1+x^2}
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\sqrt {1-x^2}
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\dfrac 12 \log (1-x^2)
Explanation
Given that,
(1-x^2) \dfrac{dy}{dx}-xy=1
Dividing both sides by
(1-x^2)
, we get
\Rightarrow \dfrac{dy}{dx}-\dfrac{x}{1-x^2}y=\dfrac{1}{1-x^2}
Which is a linear differential equation.
\therefore IF=e^{-\displaystyle \int{\dfrac{x}{1-x^2}}dx}
Put
1-x^2=t \Rightarrow -2xdx=dt \Rightarrow xdx =-\dfrac{dt}{2}
Now,
IF =e^{\dfrac 12 \displaystyle \int{\dfrac {dt}{t}}}=e^{\dfrac 12 \log x}=e^{\dfrac 12 \log (1-x^2)}=\sqrt{1-x^2}
Which of the following is a second order differential equation?
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(y')^2+x=y^2
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y'y''+y=\sin x
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y''' +(y'')^2+y=0
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y'=y^2
Explanation
The second order differential equation is
y'y"+y=\sin x
.
y=x
is a particular solution of the differential equation
\dfrac {d^2y}{dx^2}-x^2 \dfrac {dy}{dx}+xy=x
.
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True
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False
Explanation
False,
y=x
because does not satisfy the given differential equation.
x+y=\tan^{-1}y
is a solution of differential equation
y^2 \dfrac {dy}{dx}+y^2+1=0
.
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True
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False
Explanation
True,
x+y=\tan^{-1}y\Rightarrow 1+\dfrac {dy}{dx}=\dfrac {1}{1+y^2}\dfrac {dy}{dx}
\Rightarrow \ \dfrac {dy}{dx}\left(\dfrac {1}{1+y^2}-1\right)=1
, i.e.,
\dfrac {dy}{dx}=\dfrac {-(1+y^2)}{y^2}
which satisfies the given equation.
State true or false.
The solution of
\dfrac{dy}{dx}=\left(\frac{y}{x}\right)^{\tfrac{1}{3}}
is
y^{\tfrac{2}{3}}-x^{\tfrac{2}{3}}=C
.
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True
0%
False
Explanation
Given differential equation
\dfrac{dy}{dx}=\left(\dfrac{y}{x}\right)^{{1}/{3}}
\Rightarrow \dfrac{dy}{dx}=\dfrac{y^{{1} / {3}}}{x^{{1} / {3}}}
\Rightarrow y^{{-1} / {3}} dy = x^{{1} / {3}} dx
On integrating both sides, we get
\displaystyle \int y^{{-1} / {3}} dy =\int x^{{1} / {3}} dx
\Rightarrow \dfrac{y^{{-1} / {3+1}}}{\tfrac{-1}{3}+1}=\dfrac{x^{{1} / {3+1}}}{\tfrac{-1}{3}+1} + C^{'}
\Rightarrow \dfrac{2}{3}y^{2/3}=\dfrac{2}{3}x^{2/3} +C^{'}
\Rightarrow y^{2/3}-x^{2/3} =C^{'}
[where,\ \dfrac{2}{3}C^{'}=C]
The order of the differential equation
(2 x^{2} \cfrac{d^{2} y}{d x^{2}}-3 \cfrac{d y}{d x}+y=0)
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2
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1
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0
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Not defined
Explanation
Order: Order of a differential equation is the order of the highest order derivative (also known as differential coefficient) present in the equation
Here the highest order derivative is of order
2
.
hence the order is
2
.
hence the correct answer is (A)
The solution of the differential equation
\dfrac {dy}{dx}=\dfrac {1+y^2}{1+x^2}
is :
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y=\tan^{-1}x
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y-x=k(1+xy)
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x=\tan^{-1}y
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\tan ( xy)=k
Explanation
Given that,
\dfrac{1+y^2}{1+x^2}
\Rightarrow \dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}
On integrating both sides, we get
\displaystyle \int{\dfrac{dy}{1+y^2}}=\displaystyle \int{\dfrac{dx}{1+x^2}}
\tan^{-1}y=\tan^{-1}x+C
\Rightarrow \tan^{-1}y-\tan^{-1}x=C
\Rightarrow \tan^{-1}\left( \dfrac{y-x}{1+xy}\right)=C
\Rightarrow \dfrac{y-x}{1+xy}=\tan C
\Rightarrow y-x=\tan C (1+xy)
\Rightarrow y-x=K(1+xy)
Where,
k=\tan C
Which of the following is the general solution of
\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0
?
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y=(Ax+B) e^x
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y=(Ax+B) e^{-x}
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y=Ae^x +Be^{-x}
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y=A\cos x+B\sin x
Explanation
Given that,
\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0
D^2y-2Dy+y=0
,
Where,
D=\dfrac{d}{dx}
(D^2-2D+1)y=0
The auxiliary equation is
m^2-2m+1=0
(m-1)^2=0 \Rightarrow m=1, 1
Since, the roots are real and equal.
\therefore CF=(Ax+B)e^x\\ \Rightarrow y=(Ax+B)e^x
[ since, if roots of Auxiliary equation are real and equal say (m), then
CF=(C_1x+C_2)e^{mx}]
Solution of
\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\frac{dy}{dx}=0
, is:
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\log\left(\displaystyle \frac{x}{1+\sqrt{1+x^{2}}}\right)+\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=c
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\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)+\sqrt{1-x^{2}}+\sqrt{1+y^{2}}=c
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\displaystyle \log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c
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\displaystyle \log\left(\sqrt{1+x^{2}}-\sqrt{1+y^{2}}\right)+\log\left(\frac{x}{\sqrt{1+x^{2}}}\right)=c
Explanation
Given,
\displaystyle \sqrt{1+x^{2}+y^{2}+x^{2}y^{2}}+xy\dfrac{dy}{dx}=0
\Rightarrow \sqrt{1+y^{2}}\sqrt{1+x^{2}} = -xy\dfrac{dy}{dx}
Integrating both sides, we get
\Rightarrow \displaystyle\int {\dfrac{\sqrt{1+x^{2}}}{x}dx} = -\displaystyle\int{\dfrac{y}{\sqrt{1+y^{2}}}}dy
Substitute,
1+ y^{2}=t^{2}
\Rightarrow
2y dy = 2t dt
\Rightarrow
\displaystyle\int {\dfrac{\sqrt{1+x^{2}}}{x}dx} = - \displaystyle\int dt=- t+c = -\sqrt{1+y^{2}} +c
Now substitute,
1+x^{2}= k^{2}
\Rightarrow
2x dx = 2k dk
\Rightarrow \displaystyle\int\dfrac{k^{2}}{x^{2}}dk = -\sqrt{1+y^{2}} +c
\Rightarrow \displaystyle\int\dfrac{k^{2}}{k^{2}-1}dk = -\sqrt{1+y^{2}} +c
\Rightarrow \displaystyle\int\left[\dfrac{k^{2}-1}{k^{2}-1}+\dfrac{1}{k^{2}-1}\right]dk = -\sqrt{1+y^{2}} + c
\Rightarrow k +\dfrac{1}{2}\log{\dfrac{k-1}{k+1}} = -\sqrt{1+y^{2}} +C
Since ,
\displaystyle\int\dfrac{1}{x^{2}-d^{2}}dx=\dfrac{1}{2d}\log{\dfrac{x-d}{x+d}}+\mbox{constant}
\therefore \sqrt{1+x^{2}} +\dfrac{1}{2}\log{\dfrac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}}+\sqrt{1+y^{2}}= C
\Rightarrow \sqrt{1+x^{2}} +\log\left(\dfrac{x}{\sqrt{1+x^{2}}+1}\right)+\sqrt{1+y^{2}}= C
lf
f (x)
and
g (x)
are two solutions of the differential equation
a\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}
, then
f (x) - g (x)
is the solution of
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a^{2}\displaystyle \frac{d^{2}y}{dx^{2}}+\displaystyle \frac{dy}{dx}+y=e^{\displaystyle x}
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\displaystyle {a}^{2}\displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}
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a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{x}^{2}\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}=0
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a \displaystyle \frac{\mathrm{d}^{2}\mathrm{y}}{\mathrm{d}\mathrm{x}^{2}}+\mathrm{y}=\mathrm{e}^{\displaystyle \mathrm{x}}
Explanation
Given, Differential equation as
a\dfrac{\mathrm{d^2} y}{\mathrm{d} x^2}+x^2\dfrac{\mathrm{d} y}{\mathrm{d} x}+y=e^x
has
f(x),g(x)
as two solutions.
On substituting
f(x)
in the given Differential equation,
\Rightarrow af''(x)+x^2f'(x)+f(x)=e^x ................ A
On substituting
g(x)
in the given differential equation,
\Rightarrow ag''(x)+x^2g'(x)+g(x)=e^x ................B
Subtracting B from A,
\Rightarrow a\times (f''(x)-g''(x))+x^2\times(f'(x)-g'(x))+(f(x)-g(x))=0
It is in the form of
a\dfrac{\mathrm{d^2} y}{\mathrm{d} x^2}+x^2\dfrac{\mathrm{d} y}{\mathrm{d} x}+y=0
, which have a solution of
y=f(x)-g(x)
Solution of
\dfrac{d^{2}y}{dx^{2}}
=
\log x
is:
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y=\dfrac{1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}
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y=\dfrac{1}{2}x^{2}\log x+\dfrac{3}{4}x^{2}+c_{1}x+c_{2}
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y=\dfrac{-1}{2}x^{2}\log x-\dfrac{3}{4}x^{2}+c_{1}x+c_{2}
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y=3x^{2}+4x+c_{1}
Explanation
\dfrac{d^{2}y}{dx^{2}}=\log x
\Rightarrow \dfrac{d}{dx}y_{1}=\log x
\Rightarrow \int dy_{1}=\int \log x dx+c_{1}
\int \log dx=x\log x-x+{ c }_{ 1 }
\Rightarrow y_{1}=x\log x-x+c_{1}
\int dy = \int (x \log x-x+c_{1})dx+c_{2}
y=\int x\log xdx-\dfrac{x^{2}}{2}+c_{1}x+c_{2}
\int x\log xdx=x(x\log x-x)-\int(x\log x-x)dx
\rightarrow 2\int x\log xdx=x^{2}\log x-x^{2}+x^{2}/_2=x^{2}\log x^-x^{2}/_2
\Rightarrow \int x\log xdx=\dfrac{x^{2}\log x}{2}-x^{2}/_4
\Rightarrow y=\dfrac{x^{2}\log x}{2}-\dfrac {3x^{2}}{4}+c_{1}x+c_{2}
lf the solution of the differential equation
\displaystyle \frac{1}{\sin^{-1}x}(\frac{dy}{dx})=1
is
y=x \sin^{-1}x+f(x)+c
then f (x) is
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1-x^{2}
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\sqrt{1-x^{2}}
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1+x^{2}
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\sqrt{1+x^{2}}
The solution of
\displaystyle \frac{dy}{dx}+\frac{x(1+y^{3})}{y^{2}(1+x^{2})}=0
is:
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(1-x^{2})+(1+y^{3})=c
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(1+y^{2})+(1+x^{3})=c
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(1+x^{2})^{3}(1+y^{3})^{2}=c
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(1+x^{2})+(1+y^{3})=cxy^{2}
Explanation
\dfrac{dy}{dx}+\dfrac{x\left ( 1+y^{3} \right )}{y^{2}\left ( 1+x^{2} \right )}=0
\Rightarrow \dfrac{1}{3}\left ( \dfrac{3y^{2}}{1+y^{3}}dy \right )+\dfrac{1}{2}\left ( \dfrac{2x}{1+x^{2}}dx \right )=0
put
1+x^{2}=t;
1+y^{3}=k
\Rightarrow 2xdx=dt
\Rightarrow 3y^{2}dy=dk
\displaystyle\Rightarrow \dfrac{1}{3}\int \dfrac{dk}{k}+\dfrac{1}{2}\int \dfrac{dt}{t}=c
\Rightarrow \log k^{\frac{1}{3}}t^{\frac{1}{2}}=c
\Rightarrow \log k^{2}+3=log c
\left ( 1+y^{3} \right )^{2}\left ( 1+x^{2} \right )^{3}=c
A tangent drawn to the curve
\mathrm{y}=\mathrm{f}(\mathrm{x})
at
\mathrm{P}(\mathrm{x}, \mathrm{y})
cuts the
\mathrm{x}
-axis and
\mathrm{y}
-axis at A and
\mathrm{B}
respectively such that BP: AP
=3
: 1, given that
\mathrm{f}(1)=1
, then
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equation of curve is
\displaystyle \mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-3\mathrm{y}=0
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normal at (1, 1) is
\mathrm{x}+3\mathrm{y}=4
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curve passes through (2, 1/8)
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equation of curve is
\displaystyle \mathrm{x}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+3\mathrm{y}=0
Explanation
Given
BP:AP=3:1.
Then equation of tangent is
\ Y-y=f'\left( x \right) \left( X-x \right)
The intercept on the coordinate axes are
\displaystyle\ A\left( x-\frac { y }{ f'\left( x \right) } =0 \right)
and
\ B\left[ 0,y-xf'\left( x \right) \right]
Since,
P
is internally intercepts a line
AB,
\left [x=\left (\dfrac{mx_1+nx_2} {m+n} \right )\right ]
by using this formula
\displaystyle\ \therefore x=\frac { 3\left( x-\dfrac { y }{ f'\left( x \right) } \right) +1\times 0 }{ 3+1 }
\displaystyle\ \Rightarrow \frac { dy }{ dx } =\frac { y }{ -3x }
\displaystyle\ \Rightarrow \frac { dy }{ y } =-\frac { 1 }{ 3x } dx
On integrating both sides , we get
\ x{ y }^{ 3 }=C
Since, curve passes through
\\ \left( 1,1 \right) , then\quad c=1
\ \therefore x{ y }^{ 3 }=1
\displaystyle\ \therefore
At
\displaystyle x=\frac { 1 }{ 8 } \Rightarrow y=2
The solution of
(1-x^{2})\displaystyle \frac{dy}{dx}+xy=xy^{2}
is:
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y(c+\sqrt{1-x^{2}})=1-x^{2}
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y(c-\sqrt{1-x^{2}})=1-x^{2}
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y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}
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y(c-\sqrt{1-x^{2}})=\sqrt{1-x^{2}}
Explanation
\dfrac{dy}{dx}+\dfrac{xy}{1-x^{2}}=\dfrac{xy^{2}}{1-x^{2}}
\dfrac{dy}{dx}=\dfrac{xy(y-1)}{1-x^{2}}
\Rightarrow \dfrac{dy}{y(y-1)}=\dfrac{x}{1-x^{2}}dx
\Rightarrow\displaystyle \int \left(\frac{1}{y-1}-\dfrac{1}{y}\right)dy=-\dfrac{1}{2}\int \dfrac{-2x}{1-x^{2}}dx+c_{1}
\Rightarrow \log\left(\dfrac{y-1}{y}\right)=-\dfrac{1}{2}\log(1-x^{2})+\log{c_{1}}
\Rightarrow \dfrac{y-1}{y}=\dfrac{c_{1}}{\sqrt{1-x^{2}}}
\Rightarrow y\sqrt{1-x^{2}}-\sqrt{1-x^{2}}=c_{1}y
\Rightarrow y(\sqrt{1-x^{2}}-c_{1})=\sqrt{1-x^{2}}
Put
c_{1}=-c
\Rightarrow y(c+\sqrt{1-x^{2}})=\sqrt{1-x^{2}}
Solution of
\>y-x\displaystyle \frac{dy}{dx}=5\left(y^{2}+\frac{dy}{dx}\right)
, is:
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(5+x)(1-5y)=cy
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\dfrac { 5+x }{ 1-5y } =cy
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(5-x)(1+5y)=cy
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(5+x)(5-x)=c
Explanation
5y^{2} - y = -\dfrac{dy}{dx} (5+x)
\Rightarrow -\dfrac{dx}{5+x} = \dfrac{dy}{y(5y-1)}
\Rightarrow -\dfrac{dx}{5+x} = \dfrac{5y - (5y-1)}{y(5y-1)}dy
Integrating both sides, we get
\Rightarrow\displaystyle\int\dfrac{dx}{5+x} = \int\left(\dfrac{5}{1-5y} + \frac{1}{y}\right)dy
\Rightarrow \log(5+x) = -\log(1-5y)+\log y + \log c; (\log c
is a constant
)
\Rightarrow yc = (5+x)(1-5y)
Solve :
\dfrac{dy}{dx}=\dfrac{x\left ( 2\ln x+1 \right )}{\sin y+y\cos y}
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y\sin y=x^{2}\ln x+c
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y\sin x=y^{2}\ln x+c
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y\sin y=x^{2}\ln x-c
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y\sin x=y^{2}\ln x-c
Explanation
\displaystyle \frac { dy }{ dx } =\frac { x\left( 2\log { x } +1 \right) }{ \sin { y } +y\cos { y } } \Rightarrow \frac { dy }{ dx } \left( \sin { y } +y\cos { y } \right) =x+2x\log { x }
Integrating both sides
\displaystyle \int { \frac { dy }{ dx } \left( \sin { y } +y\cos { y } \right) dx } =\int { \left( x+2x\log { x } \right) dx } \\ \Rightarrow y\sin { y } ={ x }^{ 2 }\log { x } +c
If
\dfrac{dy}{dx}=xy+2x+3y+6
, then find the value of
y(-1)-e^2y(-3)
.
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e^2-1
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e^2+1
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2(e^2-1)
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-2(e^2-1)
Explanation
\dfrac{dy}{dx}=(x+3)(y+2)
, If
X=x+3, Y=y+2
\dfrac{dY}{Y}=XdX
\Rightarrow Y=A. e^{X^2/2}, y=-2 +Ae^{(x+3)^2/2}
y(-1)=-2+Ae^2, y(-3)=-2+A
y(-1)-e^2 y(-3)=-2 +2e^2 =2(e^2-1)
Solve:
y-x\dfrac{dy}{dx}=a\left ( y^{2}+\dfrac{dy}{dx} \right )
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y=c(1-ay)(x+a)
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2y=c(1-ay)(x+a)
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y=c(1-ay)(x+a)/2
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2y=c(1+ay)(x+a)
Explanation
Given,
y-\dfrac{xdy}{dx}=ay^{2}+\dfrac{a.dy}{dx}
y-ay^{2}=(x+a).\dfrac{dy}{dx}
\dfrac{dx}{x+a}=\dfrac{dy}{y(1-ay)}
\int \dfrac{dx}{x+a}=\int \dfrac{1-ay+ay}{y(1-ay)}.dy
ln(x+a)=\int \dfrac{1}{y}+\dfrac{a}{1-ay}.dy
ln(x+a)=ln(y)-ln(1-ay)+lnc
ln(x+a)=ln(\dfrac{cy}{1-ay})
(x+a)=\dfrac{cy}{1-ay}
.
\dfrac{dy}{dx}+\dfrac{\sqrt{\left ( x^{2}-1 \right )\left ( y^{2}-1 \right )}}{xy}=0
Solve the above equation:
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\sqrt{x^{2}-1}-\sec ^{-1}x+\sqrt{y^{2}-1}=c
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\sqrt{x^{2}-1}-\sec ^{-1}x+\sqrt{y^{2}-1}=-c
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\sqrt{x^{2}-1}+\sec ^{-1}x+\sqrt{y^{2}-1}=c
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\sqrt{x^{2}-1}-\sec ^{-1}x-\sqrt{y^{2}-1}=c
Explanation
\displaystyle \frac { dy }{ dx } =\frac { \sqrt { { x }^{ 2 }-1 } \sqrt { { y }^{ 2 }-1 } }{ xy } \Rightarrow \frac { y }{ \sqrt { { y }^{ 2 }-1 } }dy =\frac { \sqrt { { x }^{ 2 }-1 } }{ x } dx
\displaystyle \Rightarrow \int { \frac { y }{ \sqrt { { y }^{ 2 }-1 } } dy } =\int { \frac { \sqrt { { x }^{ 2 }-1 } }{ x } dx }
\displaystyle \Rightarrow -\sqrt { { y }^{ 2 }-1 } =\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) } +\sqrt { { x }^{ 2 }-1 } +c\\ \Rightarrow \sqrt { { x }^{ 2 }-1 } +\sqrt { { y }^{ 2 }-1 } -\sec ^{ -1 }{ x } =c
The general solution of differential equation
x\left ( 1+y^{2} \right )\mathrm{d} x+y\left ( 1+x^{2} \right )\mathrm{d} y=0
is/are:
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\displaystyle \frac{1}{2} \ln\left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=k
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\displaystyle \left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=c
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\displaystyle \left ( 1+y^{4} \right )=c\left ( 1+x^{2} \right )
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\displaystyle \ln\left [\frac{(1+x^{2})}{ (1+y^{2}) } \right ]= k
Explanation
Given equation is
x\left ( 1+y^{2} \right )\mathrm{d} x+y\left ( 1+x^{2} \right )\mathrm{d} y=0
\Rightarrow x\left ( 1+y^{2} \right )\mathrm{d} x=-y\left ( 1+x^{2} \right )\mathrm{d} y
integrating on both sides
\Rightarrow\displaystyle \int \frac{x}{1+x^{2}}\mathrm{d} x=-\int \frac{y}{1+y^{2}}\mathrm{d} y
Put
1+{ x }^{ 2 }=t \Rightarrow 2xdx=dt
and
1+y^{ 2 }=u \Rightarrow 2ydy=du
\displaystyle \frac { 1 }{ 2 } \int { \frac { dt }{ t } } =-\frac { 1 }{ 2 } \int { \frac { du }{ u } }
\Rightarrow\displaystyle \frac{1}{2}ln\left ( 1+x^{2} \right )=-\frac{1}{2}ln\left ( 1+y^{2} \right )+k
\Rightarrow ln\left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=2k
\Rightarrow \left ( 1+x^{2} \right )\left ( 1+y^{2} \right )=e^{2k}
If the length of tangent at any point on the curve
y=f(x)
intercepted between the points and the x-axis is of length
1
. Find the equation of the curve.
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\displaystyle \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =x+c
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\displaystyle \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =-x+c
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\displaystyle \sqrt { 1-{ y }^{ 2 } } +\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =x+c
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\displaystyle \sqrt { 1-{ y }^{ 2 } } +\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =-x+c
Explanation
Since, the length of tangent
\displaystyle =\left| y\sqrt { 1+{ \left( \frac { dx }{ dy } \right) }^{ 2 } } \right| =1
\displaystyle \Rightarrow { y }^{ 2 }\left( 1+{ \left( \frac { dx }{ dy } \right) }^{ 2 } \right) =1
\displaystyle\therefore \frac { dy }{ dx } =\pm \frac { y }{ \sqrt { 1-{ y }^{ 2 } } }
\displaystyle \Rightarrow \int { \frac { \sqrt { 1-{ y }^{ 2 } } }{ y } } dy=\pm \int { x } dx
\displaystyle\Rightarrow \int { \frac { \sqrt { 1-{ y }^{ 2 } } }{ y } } dy=\pm x+c
Substitute
y=\sin { \theta } \Rightarrow dy=\cos { \theta } d\theta
\displaystyle \therefore \int { \frac { \cos { \theta } }{ \sin { \theta } } } .\cos { \theta } d\theta =\pm x+c\Rightarrow \int { \frac { \cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } } } .\sin { \theta } d\theta =\pm x+c
Again substitute
\cos { \theta } =t\Rightarrow -\sin { \theta } d\theta =dt
\displaystyle \therefore \int { \frac { { t }^{ 2 } }{ 1-{ t }^{ 2 } } } dt=\pm x+c,\Rightarrow \int { \left( 1-\frac { 1 }{ 1-{ t }^{ 2 } } \right) } dt=\pm x+c
\displaystyle \Rightarrow t-\log { \left| \frac { 1+t }{ 1-t } \right| = } \pm x+c\Rightarrow \sqrt { 1-{ y }^{ 2 } } -\log { \left| \frac { 1+\sqrt { 1-{ y }^{ 2 } } }{ 1-\sqrt { 1-{ y }^{ 2 } } } \right| } =\pm x+c
The solution of differential equation
(e^x + 1) y dy = (y + 1) e^x dx
is :
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(e^x + 1) (y + 1) = ce^y
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(e^x + 1) (y + 1) = ce^{-y}
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(e^x + 1) (y + 1) = ce^{2y}
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none of the above
Explanation
The given differential equation is
(e^x + 1) y dy = (y + 1) e^x dx
\Rightarrow \displaystyle \frac{ydy}{(y+1)}=\displaystyle \frac{e^{x}}{(e^{x}+1)}dx;
Integrating both sides
\Rightarrow y- log |y + 1| = log (e^x + 1) + log k
\Rightarrow y = log |(y + 1)(e^x + 1)| + log k
\Rightarrow (y + 1)(e^x + 1) = ce^y
The solution of differential equation
(e^x + 1)y dy = (y + 1) e^x dx
is:
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(e^x + 1) (y + 1) = c . e^y
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(e^x + 1) |(y + 1)| = ce^{-y}
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(e^x + 1) (y + 1) = c
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none of these
Explanation
The given differential equation is
(e^x + 1) y . dy = (y + 1) . e^x . dx
\Rightarrow \displaystyle \frac{y.dy}{(y+1)}=\displaystyle \frac{e^{x}}{(e^{x}+1)}.dx
\Rightarrow \int\left ( 1-\displaystyle \frac{1}{y+1} \right ).dy=\int \displaystyle \frac{e^{x}}{e^{x}+1}.dx
\Rightarrow y - log |y +1| = log (e^x + 1) + log k
\Rightarrow (y+1) (e^x+1)=e^y.c
Solution of
\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y } -y \right) }{ x\left( y\log { x } -x \right) }
is:
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{ x }^{ y }=c{ y }^{ x }
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xy=c
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{ \left( xy \right) }^{ x }=c
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None of these
Explanation
\displaystyle \frac { dy }{ dx } =\frac { y\left( x\log { y-y } \right) }{ x\left( y\log { x-x } \right) }
\displaystyle \Rightarrow x\left( y\log { x-x } \right) \frac { dy }{ dx } =y\left( x\log { y-y } \right)
\displaystyle \Rightarrow \left( \log { x } -\frac { x }{ y } \right) \frac { dy }{ dx } =\log { y } -\frac { y }{ x }
\displaystyle \Rightarrow \frac { y }{ x } +\log { x } .\frac { dy }{ dx } =\log { y } +\frac { x }{ y } \frac { dy }{ dx }
\displaystyle \Rightarrow \frac { d }{ dx } \left( x\log { x } \right) =\frac { d }{ dx } \left( x\log { y } \right)
\Rightarrow y\log { y } =x\log { y } +\log { c } \Rightarrow \log { { x }^{ y } } =\log { { y }^{ x } } +\log { c } \Rightarrow { x }^{ y }=c{ y }^{ x }.
If
\int_{a}^{x} ty (t) dt = x^{2} + y(x)
then y as a function of x is:
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y = 2 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}
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y = 1 - (2 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}
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y = 2 - (1 + a^{2}) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}
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none
Explanation
Given
\int_{a}^{x} ty(t)dt = x^{2} + y(x)
.....(1)
\Rightarrow xy = 2x + \displaystyle \frac{dy}{dx}
\Rightarrow x(y - 2) = \displaystyle \frac{dy}{dx}
\Rightarrow \int xdx = \int \displaystyle \frac{dy}{y - 2}
\Rightarrow \displaystyle \frac{x^{2}}{2} = ln | y - 2 | + lnc
\Rightarrow e^{\displaystyle \frac{x^{2}}{2}} = c(y - 2)
....(2)
Now, put
x = a
in eqn (1)
\Rightarrow y = -a^{2}
So, by eqn (2)
\therefore e^{ \frac { a^{ 2 } }{ 2 } }=c(-a^{ 2 }-2)
\Rightarrow c= -\dfrac { e^{ \frac { a^{ 2 } }{ 2 } } }{ (a^{ 2 }+2) }
Put this value in (2)
\therefore e^{\displaystyle \frac{x^{2}}{2}} = - \displaystyle \frac{e^{\frac{a^{2}}{2}}}{(a^{2} + 2)} (y - 2)
\Rightarrow -y + 2 = (a^{2} + 2) e^{\displaystyle \frac{x^{2} - a^{2}}{2}}
\Rightarrow y = 2 - (2 + a^{2}) e^{\dfrac{x^{2} -a^{2}}{2}}
Through any point
\left ( x,y \right )
of a curve which passes through the origin, lines are drawn parallel to the coordiante axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of
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circles
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parabolas
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hyperbolas
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straight lines
Explanation
Let
P\left ( x,y \right )
be the point on the curve passing through the origin
O\left ( 0,0 \right )
,
and let
PN
and
PM
be the lines parallel to the
x
- and
y
-axes, respectively. If the equation of the curve is
y= y\left ( x \right )
,
the area
POM
equals
\displaystyle \int_{0}^{x}ydx
and the area
PON
equals
xy-\displaystyle \int_{0}^{x}ydx
.
Assuming that
2\left ( POM \right )= PON
, we therefore have
2\displaystyle \int_{0}^{x}ydx= xy-\displaystyle \int_{0}^{x}ydx\Rightarrow 3\displaystyle \int_{0}^{x}ydx= xy
.
Differentiating both sides of this gives
3y= x\displaystyle \frac{dy}{dx}+y\Rightarrow 2y= x\displaystyle \frac{dy}{dx}\Rightarrow \displaystyle \frac{dy}{y}=2\displaystyle \frac{dx}{x}
\Rightarrow \log \left | y \right |= 2\log \left | x \right |+C\Rightarrow y= Cx^{2}
with
C
being a constant.
This solution represents a parabola.
The solutions of
v=u\displaystyle\frac{dv}{du}+\left (\displaystyle\frac{dv}{du} \right )^{2}
where
u = y
and
v=xy
are:
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y=0
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y=-4x
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xy=cy+c^{2}
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x^{2}y=cy+c^{2}
Explanation
\displaystyle v=u\frac { dv }{ du } +{ \left( \frac { dv }{ du } \right) }^{ 2 }
Differentiating w.r.t
u
, we get
\displaystyle \frac { dv }{ du } =\frac { dv }{ du } +u\frac { { d }^{ 2 }v }{ d{ u }^{ 2 } } +2\frac { dv }{ du } \frac { { d }^{ 2 }v }{ d{ u }^{ 2 } }
\displaystyle \Rightarrow \left( u+2\frac { dv }{ du } \right) \frac { { d }^{ 2 }v }{ d{ u }^{ 2 } } =0
\displaystyle \Rightarrow \frac { dv }{ du } =c
or
\displaystyle \frac { dv }{ du } =-\frac { u }{ 2 }
Putting
\displaystyle \frac { dv }{ du } =c
, we get
v=cu+{ c }^{ 2 }\Rightarrow xy=cy+{ c }^{ 2 }
Again putting
\displaystyle \frac { dv }{ du } =-\frac { u }{ 2 }
, we get
\displaystyle v=-\frac { { u }^{ 2 } }{ 4 } \Rightarrow { y }^{ 2 }=-4xy\Rightarrow y=0
or
y=-4x
\displaystyle y-x\frac{dy}{dx}=b\left ( 1+x^{2}\frac{dy}{dx} \right ).
Solve the above differential equation.
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\displaystyle y=k\left ( y-b \right )\left ( 1+bx \right ).
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\displaystyle y=k\left ( y+b \right )\left ( 1-bx \right ).
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\displaystyle y=k\left ( y+b \right )\left ( 1+bx \right ).
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\displaystyle y=k\left ( y-b \right )\left ( 1-bx \right ).
Explanation
After rearranging the terms, we get
y-b=\dfrac{dy}{dx}(bx^{2}+x)
\dfrac{dx}{x(bx+1)}=\dfrac{dy}{y-b}
\int \dfrac{bx+1-bx}{x(bx+1)}.dx=ln(y-b)
\int \dfrac{1}{x}-\dfrac{b}{bx+1}.dx=ln(y-b)
ln(x)-ln(bx+1)+ln(c)=ln(y-b)
\dfrac{cx}{bx+1}=y-b
The solution of
\displaystyle\frac{dy}{dx}+x=xe^{(n-1)y}
is:
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\displaystyle\frac{1}{n-1}\log \left (\displaystyle\frac{e^{(n-1)y}-1}{e^{(n-1)y}} \right )=x^{2}/2+C
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e^{(n-1)y}=Ce^{(n-1)y+(n-1)x^{2/2}}+1
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\log \left (\displaystyle\frac{e^{(n-1)y}-1}{(n-1)e^{(n-1)y}} \right )=x^{2}+C
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e^{(n-1)}y=ce^{(n-1)y^{2/2+x}}+1
Explanation
Given
\displaystyle \frac { dy }{ dx } +x=x{ e }^{ y\left( n-1 \right) }
\Rightarrow \dfrac { dy }{ dx } =x\left( { e }^{ y\left( n-1 \right) }-1 \right)
\displaystyle \Rightarrow \frac { dy }{ \left( { e }^{ y\left( n-1 \right) }-1 \right) } =xdx
integrating on both sides
\displaystyle \Rightarrow \frac { 1 }{ n-1 } \int { \frac { \left( n-1 \right) }{ \left( { e }^{ y\left( n-1 \right) }-1 \right) { e }^{ y\left( n-1 \right) } } { e }^{ y\left( n-1 \right) } } dy=\frac { x^{ 2 } }{ 2 } +c
\displaystyle \Rightarrow { e }^{ y\left( n-1 \right) }=c{ e }^{ \left( n-1 \right) +\left( n-1 \right) \frac { x^{ 2 } }{ 2 } }+1
The general solution of
xy^{5}=y_{4}(y_{n}=\displaystyle\frac{d^{n}y}{dx^{n}})
is given by:
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{ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{3}+C_{3}x^{2}+C_{4}x+C_{5}
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{ y }^{ -1 }=C_{1}x^{5}+C_{2}x^{4}+C_{3}x^{2}+C_{4}x+C_{5}
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{ y }^{ -1 }=C_{1}+C_{2}x+C_{3}x^{2}+C_{4}x^{3}+C_{5}x^{4}
0%
none of these
Explanation
\displaystyle x{ y }^{ 5 }=\frac { { d }^{ 4 }y }{ d{ x }^{ 4 } } \Rightarrow \frac { { d }^{ 4 }y }{ d{ x }^{ 4 } } { y }^{ -5 }=x
\displaystyle \Rightarrow \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } \frac { { y }^{ -4 } }{ -4 } =\frac { { x }^{ 2 } }{ 2 } +c
\displaystyle \Rightarrow \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \frac { { y }^{ -3 } }{ 12 } =\frac { { x }^{ 3 } }{ 6 } +cx+d
\displaystyle \Rightarrow \frac { dy }{ dx } \frac { { y }^{ -2 } }{ -24 } =\frac { { x }^{ 4 } }{ 24 } +\frac { c{ x }^{ 2 } }{ 2 } +dx+e
\displaystyle \Rightarrow \frac { { y }^{ -1 } }{ 24 } =\frac { { x }^{ 5 } }{ 120 } +\frac { c{ x }^{ 3 } }{ 6 } +\frac { d{ x }^{ 2 } }{ 2 } +ex+f
\Rightarrow { y }^{ -1 }={ C }_{ 1 }{ x }^{ 5 }+{ C }_{ 2 }{ x }^{ 3 }+{ C }_{ 3 }{ x }^{ 2 }+{ C }_{ 4 }x+{ C }_{ 5 }
The normal to a curve at
P\left ( x,y \right )
meets the
x
-axis at
G
. If the distance of
G
from the origin is twice the abscicca of
P
, then the curve is a:
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circle
0%
hyperbola
0%
ellipse
0%
parabola
Explanation
Equation of normal at
P(x,y)
is given by,
\displaystyle Y-y=-\frac{dx}{dy}(X-x)
\displaystyle \Rightarrow G=\left( x+y\frac{dy}{dx},0\right)
Now given condition is,
\displaystyle \left|x+y\frac{dy}{dx}\right|=\left|2x\right|
\displaystyle \Rightarrow y\frac{dy}{dx}=x
or
\displaystyle y\frac{dy}{dx}=-3x
\Rightarrow \displaystyle ydy=xdx
or
ydy=-3xdx
After integrating,we get
\displaystyle \frac{y^{2}}{2}=\frac{x^{2}}{2}+c
or
\displaystyle \frac{y^{2}}{2}=-\frac{3x^{2}}{2}+c
\Rightarrow x^{2}-y^{2} = -2c
or
\displaystyle 3x^{2}+y^{2}=2c
\displaystyle \Rightarrow x^{2}-y^{2}=c_{1}
or
\displaystyle 3x^{2}+y^{2}=c_{2}
The differential equation
\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}
determines a family of circles with:
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variable radii and a fixed centre
\left ( 0,1 \right )
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variable radii and a fixed centre
\left ( 0,-1 \right )
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fixed radius
1
and variable centres along the
x
-axis
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fixed radius
1
and variable centres along the
y
-axis
Explanation
\displaystyle \frac{dy}{dx}= \displaystyle \frac{\sqrt{1-y^{2}}}{y}
\Rightarrow \displaystyle \frac{y}{\sqrt{1-y^{2}}}\: dy= dx
Integrating we have
-\sqrt{1-y^{2}}= x+C
\Rightarrow \left ( x+C \right )^{2}+y^{2}= 1
.
This represents a family of circles with variable centre
\left ( -C,0 \right )
(which lies on
x
-axis) and radius
1
The general solution of the equation
\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } }
is:
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x^3-y^3=c
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x^3+y^3=c
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x^2+y^2=c
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x^2-y^2=c
Explanation
\displaystyle \frac { dy }{ dx } =\frac { { x }^{ 2 } }{ { y }^{ 2 } } \Rightarrow { y }^{ 2 }dy={ x }^{ 2 }dx
Integrating both sides
\displaystyle \int { { y }^{ 2 }dy } =\int { { x }^{ 2 }dx } \Rightarrow \frac { { y }^{ 3 } }{ 3 } =\frac { { x }^{ 3 } }{ 3 } +\frac { c }{ 3 }
\Rightarrow { x }^{ 3 }-{ y }^{ 3 }=c
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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