Differential Equations - Class 12 Engineering Maths - Extra Questions

So (a) Order: 2 Degree: 1  

$$\implies \dfrac{d y}{d x}=\text{cos}^{-1} a$$

$$\implies d{y}=\text{cos}^{-1}a d x$$

Integrating on both sides 

$$y=x\text{cos}^{-1} a+C$$

Given $$y=1$$ when $$x=0$$

$$\implies 1=0+C\implies C=1$$

So $$y=x\text{cos}^{-1} a+1$$

Differentiating w.r.t. x, we get,

$$\dfrac{dy}{dx}=c$$

Therefore,

$$x \dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}=cx+\dfrac{a}{c}$$

$$x \dfrac{dy}{dx}+\dfrac{a}{\dfrac{dy}{dx}}=y$$

This shows that $$y=cx+\dfrac{a}{c}$$ is a solution of the given differential equation.

$$\dfrac { dy }{ dx } +Py=Q$$

where$$ P$$ and $$Q$$ are functions involving $$x$$ only.

We multiply both sides of the differential equation by the integrating factor$$ I$$ which is defined as

$${ e }^{ \int { Pdx }  }$$

$$\cos x\dfrac { dy }{ dx } +y=sinx$$ 

$$\Longrightarrow \dfrac { dy }{ dx } +y\sec x=\tan x$$

Here $$P=\sec x$$

Hence, the integrating factor $$I$$ becomes $$\displaystyle{ e }^{ \int { \sec xdx }  }={ e }^{\ln { |\sec x+\tan x| }   }= { |\sec x+\tan x| }=\sec x+\tan x  $$, because both $$secx,tanx$$ are positive in first quadrant.

$$\dfrac { dy }{ dx } +Py\quad =Q$$

where $$ P$$ and $$Q$$ are functions involving x only.

We multiply both sides of the differential equation by the integrating factor $$ I$$ which is defined as

$${ e }^{ \int { Pdx }  }$$

Here $$P=\frac { 1 }{ x } $$

Hence, the integrating factor $$I$$ $$={ e }^{ \int { \frac { 1 }{ x } dx }  }={ e }^{ \ln { x }  }=x$$

Let $$tan^{-1}x = y$$

$$\therefore\ x=tany$$

Now, differentiate wrt $$x$$

$$\therefore\ 1=\dfrac{d(tany)}{dx}$$ (differentiation of $$tant$$ is $$sec^2t$$)

$$\therefore\ 1=sec^2y\ \dfrac{dy}{dx}$$

$$\therefore\ \dfrac{dy}{dx}=\dfrac{1}{sec^2y}$$

$$\therefore\ \dfrac{dy}{dx}=\dfrac{1}{1+tan^2y}$$ ($$\because sec^2t=1+tan^2t$$)

But, $$tany = x$$

$$\therefore\ \dfrac{d(tan^{-1}x)}{dx}=\dfrac{1}{1+x^2}$$

Hence the answer is $$\dfrac{1}{1+x^2}$$

Similarly, we can write: $$v'(x) + P(x)v(x) = Q(x) $$    ...(2)

i) Using equation (1) and (2), we write $$P(x)$$ and $$Q(x)$$ in terms of $$u(x)$$ and $$v(x)$$. Subtracting (2) from (1), we get $$P(x) = \dfrac{v'(x)-u'(x)}{u(x)-v(x)}$$.

Similarly, $$Q(x) = \dfrac{v'(x)u(x) - u'(x)v(x)}{u(x)-v(x)}$$

Therefore, $$I = e^{\displaystyle 1234\int \dfrac{-(u'(x)-v'(x))}{u(x)-v(x)}dx} = e^{ln(u(x)-v(x))^{-1}} = \dfrac{1}{u(x)-v(x)}$$

Hence, the general solution is, $$\big(u(x) - v(x)\big) . \displaystyle \int \dfrac{-\big( u'(x)v(x) - v'(x)u(x) \big)}{(u(x)-v(x))^2}dx + c \,. (u(x) - v(x))$$

ii.) Given $$\alpha u(x) + \beta v(x)$$ is also a solution. 

Hence, we write,  

$$(\alpha u'(x) + \beta v'(x)) + P(x).(\alpha u(x) + \beta v(x)) = Q(x)$$.

Rearranging the terms, we get: $$\alpha(u'(x) + P(x)u(x)) + \beta(v'(x) + P(x)v(x)) = Q(x)$$

Using equation (1) and (2), we get: $$\alpha Q(x) + \beta Q(x) = Q(x)$$

We can say: $$\alpha + \beta = 1$$

iii.) Let $$c_1, \, c_2 $$ and $$ c_3$$ be the constants of integration for $$u(x), v(x) $$ and $$w(x)$$ respectively.

$$\displaystyle \left ( \frac{dy}{dx} \right )^{2}+y\left ( \frac{dy}{dx} \right )=1$$

Order = $$1,$$ Degree = $$2$$ 

(ii) $$\displaystyle _{e}\left[ \frac{dy}{dx}-\frac{d^{3}y}{dx^{3}} \right ]=ln\left [ \frac{d^{5}y}{dx^{5}}+1 \right ]$$ 

(iii) $$\left ( \left ( \dfrac{dy}{dx} \right )^{1/3}+1 \right )^{2}=\dfrac{d^{2}y}{dx^{2}}$$ 

Given equation is $$\cos^{2}x\dfrac{dy}{dx}+y=\tan x$$

$$\Rightarrow \dfrac{dy}{dx}+\sec^{2}xy=\tan x.sec^{2}x$$

The genral solution of the equation,

$$\dfrac{dy}{dx}+f(x).y=g(x)$$

is $$y=(e^{-\int f(x).dx})(\int g(x)e^{\int f(x).dx}.dx+C)$$

In the equation, $$f(x)= sec^{2}x, g(x)= \tan x - \sec^{2}x$$

The general equation is,

$$y=(e^{-\int \sec^{2}xdx})(\int \tan x. \sec^{2}x e^{\tan x}dx+C)$$

$$\int \tan x. \sec^{2}x e^{\tan x}dx= \tan x.e^{\tan x}-\int \int \sec^{2}x.e^{\tan x}$$

$$=(\tan x-1)e^{\tan x}$$

$$\therefore $$ The solution is $$y=e^{-\tan x}(e^{\tan x}(\tan x-1)+C)$$

$$y=\tan x-1+Ce^{-\tan x}$$

So, order of differential equation formed $$=2$$.

$$\dfrac{dy}{dx}=\sin (x+y)+\cos (x+y)$$

$$x+y=v\rightarrow \dfrac{dy}{dx}=\dfrac{dv}{dx}-1$$

$$\dfrac{dv}{dx}-1=\sin v+\cos v$$

$$\dfrac{dv}{1+\sin v+\cos v}=dx$$

$$\int \dfrac{dv}{1+\sin v+\cos v}=\int dx$$

$$\int \dfrac{dv}{1+\frac{1-\tan ^2\frac{v}{2}}{1+\tan ^2\frac{v}{2}}+\frac{2\tan \frac{v}{2}}{1+\tan ^2\frac{v}{2}}}=\int dx$$

$$\int \dfrac{\sec ^2\frac{v}{2}}{2\left ( 1+\tan \frac{v}{2} \right )}=\int dx$$

$$\ln \left ( 1+\tan \dfrac{v}{2} \right )=x+c$$

$$\ln \left ( 1+\tan \dfrac{x+y}{2} \right )=x+c$$

$$y=(\cos ^{ -1 } )^{ 2 }\\ \cfrac { dy }{ dx } =\cfrac { d }{ dx } [\cos ^{ -1 } x]^{ 2 }\\ \cfrac { dy }{ dx } =2\cos ^{ -1 } x\cfrac { d }{ dx } (\cos ^{ -1 } x)\\ \cfrac { dy }{ dx } =2\cos ^{ -1 } x\cfrac { (-1) }{ \sqrt { ( } 1-x^{ 2 }) } \\ \cfrac { d^{ 2 } }{ dx^{ 2 } } =\cfrac { d }{ dx } (\cfrac { dy }{ dx } )=\cfrac { d }{ dx } { [2\cos ^{ -1 } x(-1)] }{ \sqrt { 1-x^{ 2 } }  }\\ \cfrac { d^{ 2 }y }{ dx^{ 2 } } ={ -2\frac { d }{ dx }  }(\cos ^{ -1 } x)\times \cfrac { 1 }{ \sqrt { 1-x^{ 2 } }  } +2\cos ^{ -1 } x\cfrac { d }{ dx } [{ -1 }{ \sqrt { 1-x^{ 2 } }  }]\\ \cfrac { d^{ 2 }y }{ dx^{ 2 } } =(-2)\cfrac { (-1) }{ \sqrt { 1-x^{ 2 } }  } +2\cos ^{ -1 } x\cfrac { d }{ dx } [-\cfrac { 1 }{ \sqrt { 1-x^{ 2 } }  } ]\\ \cfrac { d^{ 2 }y }{ dx^{ 2 } } =\cfrac { (-2)(-1) }{ \sqrt { 1-x^{ 2 } }  } +x\cos ^{ -1 } x\cfrac { 1 }{ 2 } (1-x^{ 2 })^{ \frac{3}{2}  }.(-2x)\\ (1-x^{ 2 })\cfrac { d^{ 2 }y }{ dx^{ 2 } } =(1-x^{ 2 })[\cfrac { 2 }{ (1-x^{ 2 }) } +\cfrac { (-2x)\cos ^{ -1 } x }{ (1-x^{ 2 })^{ \frac{3}{2}  } } ]=2+\cfrac { \cos ^{ -1 } x(-2x) }{ \sqrt { 1-{ x }^{ 2 } }  } +x\cfrac { dy }{ dx } =\cfrac { 2\times \cos ^{ -1 } x }{ \sqrt { 1-x^{ 2 } }  } \\ (1-x^{ 2 })\cfrac { d^{ 2 } }{ dx^{ 2 } } -x\cfrac { dy }{ dx } -x\cfrac { dy }{ dx } -2\\ =\cfrac { 2+\cos ^{ -1 } x(-2x) }{ \sqrt { 1-x^{ 2 } }  } -\cfrac { 2\times \cos ^{ -1 } (x) }{ \sqrt { 1-x^{ 2 } }  } -2\\ =0$$

Hence proved

$$y_2(x=0)=(\cfrac{d^2y}{dx^2})_(x=0)=\cfrac{2}{(1-0^2)}+\cfrac{(-2\times0)\cos^{-1}(10)}{(1-0^2)^{\frac{3}{2}}}=2\\ \therefore y_2=0 \quad when\quad x=0$$

$${x}^{y}={e}^{x-y}$$

$$\Rightarrow y\log{x}=\left(x-y\right)$$ by taking $$\log$$ on both sides

$$\Rightarrow \dfrac{y}{x}+\log{x}\dfrac{dy}{dx}=1-\dfrac{dy}{dx}$$

$$\Rightarrow \left(\log{x}+1\right)\dfrac{dy}{dx}=1-\dfrac{y}{x}$$

The order of the differential equation $$1+\left(\dfrac{dy}{dx}\right)^2=7\left(\dfrac{d^2y}{dx^2}\right)^3$$ is $$2$$

Differentiating w.r.t. x, we get,

$$x\dfrac{dy}{dx}+y=ae^x-be^{-x}+2x$$

Differentiating w.r.t. x, we get,

$$x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}+\dfrac{dy}{dx}=ae^x+be^{-x}+2$$

$$x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}=ae^x+be^{-x}+2$$

Therefore,

$$x\dfrac{d^2y}{dx^2}+2\dfrac{dy}{dx}-xy+x^2-2=(ae^x+be^{-x}+2)-(ae^x+be^{-x}+x^2)+x^2-2=0$$

Thus, $$xy=ae^x+be^{-x}+x^2$$ is a solution of the given differential equation.

Rearranging , $$\Rightarrow \left(\dfrac{dy}{dx} \right) = \dfrac{y^2}{x^2} + \dfrac{2xy}{x^2} = \left(\dfrac{y}{x} \right)^2 + 2 \left(\dfrac{y}{x} \right) $$ ....(1)

Since the differential equation is of the form $$\dfrac{dy}{dx} = f(y/x)$$

we can use separation of variables theory

Let $$y = kx$$ substitute $$y = kx$$ in the equation  ...(i)

$$\Rightarrow \dfrac{dy}{dx} = k^2 + 2k$$ ...(2)

and, since $$y = k x \Rightarrow \dfrac{dy}{dx} = \left[k + x \left(\dfrac{dk}{dx} \right) \right]$$ ....(3)

equating (2) & (3)

$$k^2 + 2k = k + x \dfrac{dk}{dx} \Rightarrow x \dfrac{dk}{dx} = k^2 + k \Rightarrow \dfrac{dk}{k^2 + k} = \dfrac{dx}{x} $$ ...(4)

$$\Rightarrow \dfrac{dk}{k (k + 1)} = \left(\dfrac{A}{k} + \dfrac{B}{k + 1} \right) dk = \left(\dfrac{A (k + 2) + B(k)}{k (kH)} \right) dk = \left(\dfrac{A + Ak + B k}{k (kH)} \right) dk$$

$$\Rightarrow A = 1 , A + B = 0$$ (equating coefficients on LHS & RHS)

$$\Rightarrow A = 1 , B = -1$$

$$\Rightarrow \displaystyle \int \dfrac{dk}{k(k + 1)} = \int \dfrac{dk}{k} - \int \dfrac{dk}{k + 1} = \dfrac{dx}{x} \Rightarrow \log (k) - \log (k + 1) = \log x + \log c$$

$$\Rightarrow \log \left(\dfrac{k}{k + 1} \right) = \log (cx) \Rightarrow \dfrac{k}{k + 1} = cx \Rightarrow \dfrac{\left[\dfrac{(y)}{(x)} \right]}{\left[\left(\dfrac{y}{x} \right) + 1 \right]} = cx$$

when $$x = 1, y = 1 \Rightarrow \dfrac{1}{(1 + 1)} = 1(c) \Rightarrow c = \dfrac{1}{2}$$

$$\Rightarrow y = (y + x) \dfrac{x}{2}$$

Seperating terms of $$y$$ and $$x$$

$$\left(\sin{y}+y\cos{y}\right)dy=x\left(2\log{x}+1\right)dx$$

By integrating both sides, we get

$$\int{\left(\sin{y}+y\cos{y}\right)dy}=\int{x\left(2\log{x}+1\right)dx}$$

$$\int{\sin{y}dy}+\int{y\cos{y}dy}=2\int{x\log{x}dx}+\int{x dx}$$

$$-cos{y}+\int{y\cos{y}dy}=2\int{x\log{x}dx}+\dfrac{{x}^{2}}{2}$$          ......$$\left(1\right)$$

Consider $$\int{y\cos{y}dy}$$

Let $$u=y\Rightarrow du=dy$$ and $$dv=\cos{y}dy\Rightarrow v=\sin{y}$$

We have $$\int{u.dv}=uv-\int{v.du}$$

$$\int{y\cos{y}dy}=y\sin{y}-\int{\sin{y}dy}$$

$$\int{y\cos{y}dy}=y\sin{y}-\left(-\cos{y}\right)$$

$$\int{y\cos{y}dy}=y\sin{y}+\cos{y}$$                                  ......$$\left(2\right)$$

Consider $$\int{x\log{x}dx}$$

Let $$u=\log{x}\Rightarrow du=\dfrac{1}{x}dx$$ and $$dv=x.dx\Rightarrow v=\dfrac{{x}^{2}}{2}$$

We have $$\int{u.dv}=uv-\int{v.du}$$

$$\Rightarrow \int{x\log{x}dx}=\log{x}\dfrac{{x}^{2}}{2}-\int{\dfrac{{x}^{2}}{2}\times \dfrac{1}{x}dx}$$

$$\Rightarrow \int{x\log{x}dx}=\dfrac{{x}^{2}\log{x}}{2}-\int{\dfrac{x}{2}dx}$$

$$\Rightarrow \int{x\log{x}dx}=\dfrac{{x}^{2}\log{x}}{2}-\dfrac{{x}^{2}}{4}$$   ......$$\left(3\right)$$

Substituting $$\left(2\right)$$ and $$\left(3\right)$$ in $$\left(1\right)$$ we get

$$-cos{y}+\int{y\cos{y}dy}=2\int{x\log{x}dx}+\dfrac{{x}^{2}}{2}$$ 

$$-\cos{y}+y\sin{y}+\cos{y}=2\left[\dfrac{{x}^{2}\log{x}}{2}-\dfrac{{x}^{2}}{4}\right]+\dfrac{{x}^{2}}{2}+c$$ where $$c$$ is the constant of integration.

$$\Rightarrow y\sin{y}={x}^{2}\log{x}-\dfrac{{x}^{2}}{2}+\dfrac{{x}^{2}}{2}+c$$

$$\Rightarrow y\sin{y}={x}^{2}\log{x}+c$$

At $$x=1,y=\dfrac{\pi}{2}$$ we have

$$\Rightarrow \dfrac{\pi}{2}\sin{\dfrac{\pi}{2}}={1}^{2}\log{1}+c$$ where $$\log{1}=0$$

$$\Rightarrow c=\dfrac{\pi}{2}$$ where $$\sin{\dfrac{\pi}{2}}=1$$

$$\therefore y\sin{y}={x}^{2}\log{x}+\dfrac{\pi}{2}$$