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CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Maths
Differential Equations
Quiz 10
Solution of the differential equation $$\left\{ \dfrac { 1 }{ x } -\dfrac { { y }^{ 2 } }{ (x-y)^{ 2 } } \right\} dy+\left\{ \dfrac { { y }^{ 2 } }{ (x-y)^{ 2 } } -\dfrac { 1 }{ y } \right\} dx=0$$ is
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$$\ln|\dfrac{x}{y}|+\dfrac{xy}{x-y}=c$$
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$$\dfrac{xy}{x-y}=ce^{x/y}$$
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$$\ln|xy|=c+\dfrac{xy}{x-y}$$
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$$none\ of\ these$$
The solution of the differential equation $$x^2\dfrac{dy}{dx}\cos\left(\dfrac{1}{x}\right) - y \sin\left(\dfrac{1}{x}\right) = -1 ; where \ y \rightarrow -1 \ as \ x \rightarrow \infty$$ is
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$$y = \sin\left(\dfrac{1}{x}\right) - \cos\left(\dfrac{1}{x}\right) $$
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$$y = \dfrac{x+1}{x\sin\left(\dfrac{1}{x}\right)}$$
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$$y = \cos\left(\dfrac{1}{x}\right) + \sin\left(\dfrac{1}{x}\right) $$
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$$y = \dfrac{x+1}{x\cos\left(\dfrac{1}{x}\right)}$$
Explanation
$$\displaystyle x^{2}\frac{dy}{dx}cos\left ( \frac{1}{x} \right )- ysin\left ( \frac{1}{x} \right )=-1$$
$$\displaystyle \frac{dy}{dx}-\frac{y}{x^{2}}tan\left ( \frac{1}{x} \right )= -\frac{1}{x^{2}}cos \left ( \frac{1}{x} \right )$$
$$\displaystyle \frac{dy}{dx}+\left ( \dfrac{-1}{x^{2}}tan\left ( \dfrac{1}{x} \right ) \right )y=\frac{-1}{x^{2}cos\left ( \dfrac{1}{x} \right )}$$
$$I.F = e^{\displaystyle\int -\frac{1}{x^{2}}tan\left ( \dfrac{1}{x} \right )dx}$$
$$\displaystyle \frac{1}{x}=t$$
$$\displaystyle -\frac{dx}{x^{2}}=dt$$
$$ =e^{\displaystyle\int tan(t)dt}$$
$$\displaystyle =e^{ln(sect)}$$
$$\displaystyle =sec\left ( \dfrac{1}{x} \right )$$
$$\displaystyle sec\left ( \frac{1}{x} \right ).y=\int \frac{1}{x^{2}cos\left ( \dfrac{1}{x} \right )}.sec\left ( \frac{1}{x} \right )$$
$$\displaystyle =-\int \frac{sec^{2}\left ( \dfrac{1}{x} \right )}{x^{2}}$$
Let $$\displaystyle \frac{1}{x}=t$$
$$\displaystyle \frac{-dx}{x^{2}}=dt$$
$$\displaystyle =\int sec^{2}(t)dt$$
$$\displaystyle \boxed{sec\left ( \frac{1}{x} \right ).y=tan\left ( \frac{1}{x} \right )+C}$$
$$\displaystyle 1.-1= 0+C$$ $$C=-1$$
$$\displaystyle sec\left ( \frac{1}{x} \right ).y= tan\left ( \frac{1}{x} \right )-1, $$
$$\displaystyle \boxed{y=sin\left ( \frac{1}{x} \right )-cos\left ( \frac{1}{x} \right )}$$
The solution of the differential equation $$\dfrac{dy}{dx} = \dfrac{x^2 + y^2 + 1}{2xy}$$ satisfying $$y (1) = 1$$, is
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a hyperbola
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a circle
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$$y^2 = x (1 + x) - 10$$
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$$(x - 2)^2 + (y - 3)^2 = 5$$
Explanation
$$\dfrac{dy}{dx}=\dfrac{x^2+y^2+1}{2xy}$$
$$2y\dfrac{dy}{dx}=x+\dfrac{y^2}{x}+\dfrac{1}{x}$$
$$2y\dfrac{dy}{dx}-\dfrac{y^2}{x}=x+\dfrac{1}{x}$$
$$y^2=u=2y\dfrac{dy}{dx}=\dfrac{du}{dx}$$
$$\dfrac{du}{dx}-\dfrac{u}{x}=x+\dfrac{1}{x}$$
$$\dfrac{du}{dx}+Pu=Q$$
$$\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=1+\dfrac{1}{x^2}$$
$$\int d\left ( \dfrac{u}{x} \right )=\int \left ( 1+ \dfrac{1}{x^2}\right )dx\Rightarrow \dfrac{u}{x} x-x^{-1}+C$$
$$\Rightarrow \dfrac{y^2}{x}=x-\dfrac{1}{x}+C$$
$$\Rightarrow y^2=x^{2}-1+x$$
The general solution of differential equation $$\dfrac{dy}{dx}=\sin^{3}{x}\cos^{2}{x}+xe^{x}$$
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$$y=\dfrac{1}{5}\cos^{5}{x}+\dfrac{1}{3}csc^{3}{x}+(x+1)e^{x}+c$$
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$$y=\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}+(x-1)e^{x}+c$$
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$$y=-\dfrac{1}{5}\cos^{5}{x}-\dfrac{1}{3}csc^{3}{x}-(x-1)e^{x}-c$$
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None of these
The value of $$\displaystyle \lim_{x \rightarrow \infty}$$ y(x) obtained from the differential equation $$\dfrac{dy}{dx} = y - y^2$$, where y(0) = 2 is
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1
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-1
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0
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$$\dfrac{2}{2-e}$$
Explanation
$$\dfrac{dy}{dx}=y-y^{2}$$ $$y(0)=2$$
$$\dfrac{dy}{y-y^{2}}=dx$$
$$\dfrac{-dy}{y^{2}-y}=dx$$
$$\dfrac{-dy}{y^{2}\left(\dfrac{1}{2}\right)^{2}-2\times \dfrac{1}{2}y-\dfrac{1}{4}}=dx$$
$$\dfrac{dy}{\left(y-\dfrac{1}{2}\right)^{2}-\left(\dfrac{1}{2}\right)^{2}}=-dx$$
$$\dfrac{1}{2\times \dfrac{1}{2}}\ln\left|\dfrac{y-\dfrac{1}{2}-\dfrac{1}{2}}{y-\dfrac{1}{2}+\dfrac{1}{2}}\right|=-x+c$$
$$\ln\left(\dfrac{y-1}{y}\right)=x+c$$
Given $$y(0)=2$$
$$\ln\left(\dfrac{2-1}{2}\right)=C$$
$$\boxed{C=\ln\left(\dfrac{1}{2}\right)}$$
$$\ln\left(\dfrac{y-1}{y}\right)=-x+\ln\left(\dfrac{1}{2}\right)$$
$$\ln\left(\dfrac{y}{y-1}\right)=x+\ln(2)$$
$$\dfrac{y}{y-1}=e^{x+\ln (2)}$$
$$\dfrac{y-1}{y}=\dfrac{1}{e^{x+\ln(2)}}$$
$$1-\dfrac{1}{y}=e^{-(x+\ln(2))}$$
$$\dfrac{1}{y}=1-e^{-(x+\ln (2)]}$$
$$y=\dfrac{1}{1-e^{-[x+\ln (2)]}}$$
$$\displaystyle\lim_{x\rightarrow \infty}\dfrac{1}{1-e^{-\infty}}=\dfrac{1}{1-0}=1$$
The solution of the differential equation $$\dfrac{dy}{dx}+\dfrac{1}{x}\tan y=\dfrac{\tan y \sin y}{{x}^{2}}$$ is
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$$\dfrac{1}{x}\cos y = \dfrac{1}{{2x}^{2}}+K$$
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$$\dfrac{1}{x}\cot y = \dfrac{1}{{2x}^{2}}+K$$
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$$\dfrac{1}{x}\csc y = \dfrac{1}{{2x}^{2}}+K$$
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$$\dfrac{1}{x}\cos x = \dfrac{1}{{2x}^{2}}+K$$
The solution of the differential equation $$(3xy+y^{2})dx+(x^{2}+xy)dy=0$$ is
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$$x^2(2x+y)=3$$
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$$y^2(2x+y)=3$$
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$$x^2y(2x+y)=3$$
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$$x^2y(2x+3)=9$$
The solution of the differential equation $$\left ({x}^{2}+1\right)\dfrac {dy}{dx}+{y}^{2}+1=0,$$ is $$\left[ ify\left( 0 \right) =1 \right]$$
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$$y=2+{x}^{2}$$
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$$y=\dfrac { \left( 1+x \right) }{ \left( 1-x \right) }$$
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$$y=x\left (x-1\right)$$
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$$y=\dfrac { \left( 1-x \right) }{ \left( 1+x \right) }$$
The solution of differential equation $$\dfrac{dy}{dx} =\dfrac{x(2\ln{x+1})}{\sin{y}+y\cos{y}}$$ is
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$$y\sin{y}=x\ln{x}+c$$
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$$y\sin{y}=x^{2}\ln{x}+c$$
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$$\sin{y}=x^{2}\ln{x}+c$$
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$$y\cos{y}=x^{2}\ln{x}+c$$
Explanation
Equation of family of circle is
$$(n+a)^2+(y-a)^2=a^2$$ or
$$x^2+y^2+2ax-2ay +a^2$$
$$=0$$ ------(1)
Diffentiating , we get
$$2n+2y \dfrac{dy}{dx}+2a-2a \dfrac{dy}{dx}=0$$
$$\Rightarrow x+y \dfrac{dy}{dx}=a\left(\dfrac{dy}{dx}-1\right)$$
$$\Rightarrow a=\dfrac{x+yy^1}{y^1-1}$$
where $$y^1=\dfrac{dy}{dx}$$
Substituting the value of a in eq $$(1)$$ and Simplifying
$$\Rightarrow (xy^1-x+x+yy^1)+(yy^1-y-x-yy^1)^2$$
$$=(x+(yy^1))^2$$
$$\Rightarrow (x+y)^2[(y^1)^2+1]=(x+yy^1)^2$$
The given differential equation is :
$$\dfrac{dy}{dx}=\dfrac{x(2\log x+1)}{\sim y +y \cos y}$$
$$\Rightarrow (\sin y+y \cos y)dy =x(2\log x+1)dx$$
$$\Rightarrow \int (\sin y +y \cos y )dy=\int x(2\log x+1)dx$$ integrating both sides
$$\Rightarrow -\cos y +y \sin y - \int \sin y \, dy$$
$$=2\log ^x \dfrac{x^2}{2}-2 \int \dfrac{1}{x} \dfrac{x^2}{2}dx +\dfrac{x^2}{2}$$
$$\Rightarrow -\cos y + y \sin y + \cos y =x^2 \log x- \dfrac{x^2}{2}+\dfrac{x^2}{2}+c$$
$$\Rightarrow y \sin y =x^2 \log x +c$$ .......(2)
It is given that when $$y=\dfrac{\pi}{2},x=1$$
So, putting $$y=\dfrac{\pi}{2}, x=1$$ in eq (2),
$$\Rightarrow \dfrac{\pi}{2} \sin \dfrac{\pi }{2}=1.\log 1+c$$
$$\Rightarrow c=\dfrac{\pi}{2}$$
Putting $$c=\dfrac{\pi}{2}$$ in eq (2) we oftain
$$\boxed{y \sin y =x^2\log x+\dfrac{\pi}{2}}$$
The equation of a curve passing through the point (0, 0) and whose differential equation is $${y'} = {e^x}\sin x$$ is
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$$2y - 1 = {e^x}\left( {\sin x - \cos x} \right)$$
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$$2y - 1 = {e^{ - x}}\left( {\cos x - \sin x} \right)$$
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$$2y + 1 = {e^{ - x}}\left( {\cos x - \sin x} \right)$$
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$$None\,of\,these$$
Explanation
given, $$\dfrac {dy}{dx}=e^x\sin x$$
$$dy=e^x\sin x.dx$$
$$\displaystyle \int dy=\displaystyle \int e^x\sin x.dx\ \rightarrow $$ integrating both sides.
$$det,\ \displaystyle \int e^x\sin x\ dx=I----(1)$$
$$y=\displaystyle \int \underset{f(x)}{\underset{\downarrow}{e^x}}\underset{g(x)}{\underset{\downarrow }{\sin x}}\ dx=I$$
Since, $$\displaystyle \int f(x)\,g(x)dx =f(x).\displaystyle \int g(x)dx-\displaystyle \int \left [f'(x) \displaystyle \int g(x)dx\right]dx$$
$$\therefore \ \Rightarrow \ e^x.\displaystyle \int \sin x\ dx-\displaystyle \int \left [e^x.\displaystyle \int \sin x\ dx\right]$$
$$I=e^x.(-\cos x)-\displaystyle \int e^x.(-\cos x).dx$$
$$I=-e^x\cos x+\displaystyle \int \underset{f(x)}{\underset{\downarrow}{e^x}}.\underset{g(x)}{\underset{\downarrow}{\cos x}}\ dx$$
Now apply same formula as above
$$I=-e^x\cos x.+\left [e^x.\displaystyle \int \cos x.dx-\displaystyle \int (ex.\displaystyle \int \cos x.dx)dx\right]$$
$$I=-{ e }^{ x }.\cos { x } +\left[ { e }^{ x }.\sin { x } -\underbrace { \displaystyle \int { { e }^{ x }.\sin { x } .dx } }_{ \left( I \right) } \right] $$
$$I=-e^x\cos x+e^x\sin x-I+c$$
$$2y=e^x(\sin x-\cos x)+c$$
Since this curve passes through $$(0,0)$$ therefore
$$2\times 0=e^0(\sin 0\cos 0)+c$$
$$c=1$$
$$\therefore \ $$ equation of curve $$2y=e^x(\sin x-\cos x)+1$$
$$(2y-1)=e^x(\sin x-\cos x)$$
Solution of the differential equation $$ye^{x/y}dx=(xe^{\frac {x}{y}}+y^{2})dy(y \neq 0)$$ is ?
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$$e^{x/y}=x+C$$
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$$e^{x/y}+x=C$$
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$$e^{y/x}=x+C$$
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$$e^{x/y}=y+C$$
Solution of differential equation $$xdy=(y-{x}^{2}-{y}^{2})dx$$ is (where $$c$$ is arbitrary constant)
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$$y=x\cos{(c+x)}$$
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$$y=x\tan{(x-c)}$$
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$$y=x\csc{(c+x)}$$
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none of these
Let $$y=y(x)$$ be the solution of the differential equation $$(1-x^2)\dfrac{dy}{dx}-xy=1,x\in(-1,1)$$ if $$y(0)=0$$, then $$y\left(\dfrac{1}{2}\right)$$ is equal to
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$$\dfrac{\pi}{3\sqrt3}$$
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$$\dfrac{\pi}{\sqrt3}$$
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$$\dfrac{\pi}{6}$$
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$$\dfrac{\pi}{3}$$
If the general solution of the differential equation if $$y'=\frac { y }{ x } +\Phi \left( \frac { x }{ y } \right) $$, for some function $$\Phi $$, is given by $$yin|cx|=x$$, where c is an arbitrary constant, then $$\Phi $$ (2) is equal to :
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4
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$$\frac { 1 }{ 4 } $$
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-4
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$$-\frac { 1 }{ 4 } $$
$$(2y+xy^{3})dx+(x+x^{2}y^{2})dy=0$$ solution of differential equation is
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$$3x^{2}y+x^{3}y^{2}=c$$
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$$3x^{2}y^{2}+xy^{2}=c$$
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$$yx^{2}+\dfrac{xy}{3}=c$$
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$$x^{2}y+\dfrac{x^{3}y^{3}}{3}=c$$
Solution of differential equation $${ x }^{ 6 }dy+3{ x }^{ 5 }ydx=xdy-2y\ dx$$ is
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$${ x }^{ 3 }y=\dfrac { y }{ { x }^{ 2 } } +C$$
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$${ x }^{ 3 }y=\dfrac { 2y }{ { x }^{ 2 } } +C$$
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$${ x }^{ 3 }{ y }^{ 2 }=\dfrac { y }{ { x }^{ 2 } } +C$$
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$${ x }^{ 3 }=\dfrac { y }{ { x }^{ 2 } } +C$$
The solution of equation $$\dfrac { dy }{ dx } =\dfrac { 1 }{ x+y+1 } $$
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$$x={ ce }^{ y }-y-2$$
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$$y=x+{ ce }^{ y }-2$$
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$$x+{ ce }^{ y }-2-2=0$$
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$$y=x$$
Let $$y=y(x)$$ be the solution of the differential equation $$\sin x\dfrac {dy}{dx}+y\cos x=4x,x\ \in\ (0,\pi)$$, if $$y\left(\dfrac {\pi}{2}\right)=0$$, then $$y\left(\dfrac {\pi}{6}\right)$$ is equal to
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$$-\dfrac {4}{9}\pi^{2}$$
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$$\dfrac {4}{9\sqrt {2}}\pi^{2}$$
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$$\dfrac {-8}{9\sqrt {2}}\pi^{2}$$
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$$-\dfrac {8}{9}\pi^{2}$$
Solution of the differential equation $$dr=a\left( r\sin\theta d\theta-\cos\theta dr \right )$$ is
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$$r\left( 1+a\cos { \theta } \right) =c$$
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$$r\left( 1+a\cos { \theta } \right) =ac$$
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$$r\left( 1-a\cos { \theta } \right) =c$$
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$$r\left( 1-a\cos { \theta } \right) =ac$$
The solutions of the differential equation $$\frac{dy}{dx}= \frac{siny+x}{sin2y-x cos y }$$ is
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$$sin^{2}y = xsin y +\frac{x^{2}}{2}+c$$
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$$sin^{2}y = xsin y -\frac{x^{2}}{2}+c$$
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$$sin^{2}y = x+sin y +\frac{x^{2}}{2}+c$$
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$$sin^{2}y = x-sin y +\frac{x^{2}}{2}+c$$
The general solution of the differential equation $$\dfrac{dy}{dx} + \sin \dfrac{x+y}{2} = \sin \dfrac{x-y}{2}$$ is
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$$\log \left(\tan \dfrac{y}{2}\right) = c-2\sin x$$
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$$\log\left( \tan \dfrac{y}{4}\right) = c-2\sin \left(\dfrac{x}{2}\right)$$
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$$\log \left(\tan \left(\dfrac{y}{2} +\dfrac{\pi}{4}\right) \right)= c-2 \sin x$$
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None of the above
If $$y\left(x\right)$$ is the solution of the differential equation $$\left(x+2\right)\dfrac{dy}{dx}=x^{2}+4x-9,x\neq -2$$ and $$y\left(0\right)=0$$, then $$y\left(-4\right)$$ is equal to
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
The solution of differential equation,$$\left(x+\tan y\right)dy=\sin 2y\,dx$$ is
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$$xcoty=\dfrac{1}{2}\log |cosec2y-cot2y|+c$$
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$$x=\tan y+c\sqrt{\tan y}$$
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$$x=\cot y+c\sqrt{\cot y}$$
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$$None\ of\ these$$
If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy)dx=xdy$$, then $$f\left(-\dfrac {1}{2}\right)$$ is equal to
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$$-\dfrac {4}{5}$$
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$$\dfrac {2}{5}$$
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$$\dfrac {4}{5}$$
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$$-\dfrac {2}{5}$$
General solution of differential equation $$x^{2}(x+y\frac{dy}{dx})+(x\frac{dy}{dx}-y)\sqrt{x^{2}+y^{2}}=0$$ is
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$$\frac{1}{\sqrt{x^{2}+y^{2}}}+\frac{y}{x}=c$$
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$$\sqrt{x^{2}+y^{2}}-\frac{y}{x}=c$$
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$$\sqrt{x^{2}+y^{2}}+\frac{y}{x}=c$$
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$$2\sqrt{x^{2}+y^{2}}+\frac{y}{x}=c$$
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$$\frac{1}{\sqrt{x^{2}+y^{2}}}-\frac{y}{x}=c$$
Solution of the differential equation $$\dfrac{dy}{dx}=\dfrac{y(1+x)}{x(y-1)}$$ is
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$$\log|xy|+x-y=C$$
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$$\log|xy|-x+y=C$$
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$$\log|xy|+x+y=C$$
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$$\log|xy|-x-y=C$$
Solution of $$(x+y-1)dx+(2x+2y-3)dy=0$$ is
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$$y+x+log(x+y-2)=c$$
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$$y+2x+log(x+y-2)=c$$
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$$2y+x+log(x+y-2)=c$$
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$$2y+2x+log(x+y-2)=c$$
The solution of $$\dfrac{xdy}{x^{2}+y^{2}}=\left(\dfrac{y}{x^{2}+y^{2}}-1\right)dx$$ is
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$$y=x\cot(c-x)$$
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$$\cos^{-1}u/x=-x+C$$
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$$y=x\tan (c-x)$$
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$$y^{2}/x^{2}=x\tan (c-x)$$
General solution of differential equation $$x^2dy + y(x + y)dx = 0$$ is
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$$2x + y = Ax^2Y$$
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$$2x - y = \dfrac{Ax^2}{y}$$
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$$2x + y = \dfrac{Ax^2}{y}$$
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$$x - 2y = Ax^2y$$
If $$y(t)$$ is solution of $$(t + 1) \dfrac{dy}{dt} - ty = 1, y(0) = -1$$, then $$y(1) =$$
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$$\dfrac{1}{4}$$
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$$-2$$
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$$-\dfrac{1}{2}$$
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$$\dfrac{1}{2}$$
Solution of the differential equation $$\cos{\,}x {\,}dy=y(\sin{\,}x-y)dx,0<x<\dfrac{\pi}{2}$$ is
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y sec x$$=$$ tan x $$+$$ c
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y sec x$$=$$ sec x $$+$$ c
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y sec x$$=$$ (sec x $$+$$ c) y
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sec x$$=$$ y (tan x $$+$$ c)
The solution of the differential equation $$ydx-xdy+3x^2y^2e^{x^3} dx=0$$ is
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$$\dfrac{x}{y}+e^{x^3}=c$$
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$$\dfrac{x}{y}-e^x=c$$
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$$\dfrac{x}{y}+ex^{x^3}-x^2y=c$$
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$$\dfrac{x}{y}-e^{x^3}=c$$
Solution of the equation $$(x+y)^2\dfrac{dy}{dx}=4,y(0)=0$$ is
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$$y=2tan^{-1}\left(\dfrac{x+y}{2}\right)$$
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$$y=4tan^{-1}\left(\dfrac{x+y}{4}\right)$$
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$$y=4tan^{-1}\left(\dfrac{x+y}{2}\right)$$
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$$y=2tan^{-1}\left(\dfrac{x+y}{4}\right)$$
The solution of the differential equation $$\log\left(\dfrac {dy}{dx}\right)=4x-2y-2,y=1$$ when $$x=1$$ is
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$$2e^{2y+2}=e^{4x}+e^{2}$$
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$$2e^{2y-2}=e^{4x}+e^{2}$$
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$$2e^{2y+2}=e^{4x}+e^{4}$$
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$$3e^{2y+2}=e^{3x}+e^{2}$$
the solution of $$ydx-xdy+3{ x }^{ 2 }{ y }^{ 2 }{ e }^{ { x }^{ 3 } }dx=0$$ is
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$$\frac { x }{ y } +{ e }^{ { x }^{ 3 } }$$=C
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$$\frac { x }{ y } -{ e }^{ { x }^{ 3 } }$$=0
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$$-\frac { x }{ y } +{ e }^{ { x }^{ 3 } }$$=C
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none of these
The solution of the differential equation $$\left(x+2y^{3}\right)\dfrac{dy}{dx}=y$$ is
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$$\dfrac{x}{y^{2}}=y+C$$
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$$\dfrac{x}{y}=y^{2}+C$$
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$$\dfrac{x^{2}}{y}=y^{2}+C$$
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$$\dfrac{x}{y}=x^{2}+C$$
General solution of the differential equation (x+y-2) dy=(x+y) dx is
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y+x=log(y-x+1)+c
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y-x log (x+y-1)+c
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y-2x=log (x+y-1)+c
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y+2x=log (x+y+1)+c
The solution of $$x^2dy-y^2dx+xy^2(x-y)dy=0$$ is?
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$$ln\left|\dfrac{x-y}{xy}\right|=\dfrac{y^2}{2}+c$$
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$$ln\left|\dfrac{xy}{x-y}\right|=\dfrac{x^2}{2}+c$$
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$$ln\left|\dfrac{x-y}{xy}\right|=\dfrac{x^2}{2}+c$$
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$$ln\left|\dfrac{x-y}{xy}\right|=x+c$$
Solution of $$\left(1+xy\right)ydx+\left(1-xy\right)xdy=0$$ is
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$$\log\dfrac{x}{y}+\dfrac{1}{xy}=c$$
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$$\log\dfrac{x}{y}=c$$
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$$\log\dfrac{x}{y}-\dfrac{1}{xy}=c$$
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$$\log\dfrac{y}{x}-\dfrac{1}{xy}=c$$
The general solution of the differential equation $$y-x\dfrac{dy}{dx}=y^{2}\cos x\left(1-\sin x\right)$$ is
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$$y\left(\sin x+c\right)=\tan x+\sec x$$
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$$y\left(\sin x+c\right)=\tan x-\sec x$$
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$$\dfrac{x}{y}=\sin x+\dfrac{1}{4}\cos 2x+c$$
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$$\dfrac{x}{y}=\sin x+\dfrac{1}{8}\cos 2x+c$$
Solution of $$\dfrac { d y } { d x } + 2 x y = y$$ is
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$$y = c e ^ { x - x ^ { 2 } }$$
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$$y = c e ^ { x ^ { 2 } } - x$$
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$$y = c e ^ { x }$$
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$$y = c e ^ { - x ^ { 2 } }$$
Let =y(x) be the solution of the differential equation, x$$\frac { dy }{ dx } +y=x{ log }_{ e }x,(x>1).$$ If $$2y(2)={ log }_{ e }4-1,$$ then y(e) is equal to :
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$$-\dfrac { { e }^{ 2 } }{ 2 } $$
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$$\dfrac { e } { 4 } $$
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$$\dfrac { { e }^{ 2 } }{ 4 } $$
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$$-\dfrac { { e }^{ 2 } }{ 4} $$
The solution of the Differential Equation $$(x+y)(dx-dy)=dx+dy$$ is
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$$l\ n(x-y)=x+y+C$$
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$$l\ n(x+y)=x-y+C$$
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$$l\ n(x-y)=x-y-C$$
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$$none\ of\ these$$
The integrating factor of the equation $$y^{2}dx+(3xy-1)dy=0$$ is
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$$y^{2}$$
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$$y^{3}$$
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$$\dfrac {1}{y^{2}}$$
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$$\dfrac {1}{y^{3}}$$
The solution of the differential equation $$\dfrac{dy}{dx}=\sec (x+y)$$ is
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0%
$$y-\tan\dfrac{x+y}{2}=c$$
0%
$$y+\tan\dfrac{x+y}{2}=c$$
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$$y+2\tan\dfrac{x+y}{2}=c$$
0%
$$None\ of\ these$$
The solution of differential equation $$dy/dx -y/x+\tan y/x$$ is:
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0%
$$x=C'\cos (y/x)$$
0%
$$x=C'\sin (y/x)$$
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$$x=C'\csc (y/x)$$
0%
$$none\ of\ these$$
The solution of the equation $$\dfrac{dy}{dx}=\dfrac{(1+x)y}{(y-1)x}$$ is:
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0%
$$\log xy+x+y=c$$
0%
$$\log (\dfrac{x}{y})+x-y=c$$
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$$y-x-\log xy=c$$
0%
$$None\ of\ these$$
Solution of $$\left(\dfrac {dy}{dx}\right)^{2}+x\dfrac {dy}{dx}-y=0$$ is
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0%
$$y=3x^{2}+9$$
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$$y=3x+9$$
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$$y=\dfrac {4}{3}x^{2}$$
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$$y=9x+3$$
Number of values of m $$\in$$ N, for which $$y=e^{mx}$$ is a solution of the differential equation $$D^3y-3D^2y-4Dy+12y=0$$, is?
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0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
More than $$2$$
The solution the differential equation $$\left(\dfrac{dy}{dx}\right)^{2}-\dfrac{dt}{dx}(e^{x}+e^{-x})+1=0$$ is/are
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0%
$$y+e^{-x}=c$$
0%
$$y-e^{-x}=c$$
0%
$$y+e^{x}=c$$
0%
$$y-e^{x}=c$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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