Question

If $ \dfrac{dy}{dx}+x\sin 2y={{x}^{3}}co{{s}^{2}}y,y(0)=0 $ , then $ y(1) $ is equal to:
A. $ {{\tan }^{-1}}e $
B. $ {{\tan }^{-1}}\dfrac{1}{e} $
C. $ {{\tan }^{-1}}\dfrac{e}{2} $
D. $ {{\tan }^{-1}}\dfrac{1}{2e} $

Answer 1

Hint: Steps to solve a First Order Linear Differential Equation:
Convert into the standard form $ \dfrac{dy}{dx}+P\times y=Q $ , where P and Q are constants or functions of x only.
Find the Integrating Factor (F) by using the formula: $ F={{e}^{\int{P}\;dx}} $ .
Write the solution using the formula: $ y\times F=\int{Q\times F\ dx}+C $ where C is the constant of integration.
Find the value of C by using the values of $ y(0) $ , $ y(1) $ etc.
Make an appropriate substitution to get rid of the dependent variable (y) from all the other terms of the given differential equation.
Simplify the solution obtained using an appropriate substitution and integrating by parts

Complete step-by-step answer:
The given differential equation $ \dfrac{dy}{dx}+x\sin 2y={{x}^{3}}co{{s}^{2}}y $ is not in the standard form yet.
We divide both sides by $ {{\cos }^{2}}y $ and expand $ \sin 2y=2\sin y\cos y $ , to get:
⇒ $ \dfrac{1}{co{{s}^{2}}y}\dfrac{dy}{dx}+\dfrac{x\times 2\sin y\cos y}{co{{s}^{2}}y}={{x}^{3}} $
⇒ $ {{\sec }^{2}}y\dfrac{dy}{dx}+2x\tan y={{x}^{3}} $
Substituting $ \tan y=z $ , we have $ {{\sec }^{2}}ydy=dz $ .
⇒ $ \dfrac{dz}{dx}+2x\times z={{x}^{3}} $
Comparing the above differential equation with the standard form $ \dfrac{dy}{dx}+P\times y=Q $ , we can say that:
 $ P=2x $ and $ Q={{x}^{3}} $ .
∴ The solution is:
⇒ $ z\times {{e}^{\int{2x}\;dx}}=\int{{{x}^{3}}\times {{e}^{\int{2x}\;dx}}\ dx}+C $
Using $ \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1} $ , we get:
⇒ $ z\times {{e}^{2\times \dfrac{{{x}^{2}}}{2}}}=\int{{{x}^{3}}\times {{e}^{2\times \dfrac{{{x}^{2}}}{2}}}\ dx}+C $
⇒ $ z\times {{e}^{{{x}^{2}}}}=\int{{{x}^{2}}\times x\times {{e}^{{{x}^{2}}}}\ dx}+C $
Let's substitute $ {{x}^{2}}=t $ , then $ 2xdx=dt $ .
⇒ $ z\times {{e}^{t}}=\dfrac{1}{2}\int{t{{e}^{t}}\ dt}+C $
Integrating by parts, this is equal to:
⇒ $ z\times {{e}^{t}}=\dfrac{1}{2}\int{t{{e}^{t}}\ dt}+C $
⇒ $ z\times {{e}^{t}}=\dfrac{1}{2}\left[ t\int{{{e}^{t}}\ dt}-\int{1\times \left( \int{{{e}^{t}}dt} \right)dt} \right]+C $
⇒ $ z\times {{e}^{t}}=\dfrac{1}{2}\left( t{{e}^{t}}-{{e}^{t}} \right)+C $
Dividing by $ {{e}^{t}} $ , we get:
⇒ $ z=\dfrac{1}{2}\left( t-1 \right)+\dfrac{C}{{{e}^{t}}} $
Back substituting for $ t={{x}^{2}} $ and $ z=\tan y $ , we get:
⇒ $ \tan y=\dfrac{1}{2}\left( {{x}^{2}}-1 \right)+\dfrac{C}{{{e}^{{{x}^{2}}}}} $
⇒ $ y={{\tan }^{-1}}\left[ \dfrac{1}{2}\left( {{x}^{2}}-1 \right)+\dfrac{C}{{{e}^{{{x}^{2}}}}} \right] $
We are given that $ y(0)=0 $ . Putting $ x=0 $ and equating it to 0, we get:
⇒ $ 0={{\tan }^{-1}}\left[ \dfrac{1}{2}\left( 0-1 \right)+\dfrac{C}{{{e}^{0}}} \right] $
⇒ $ \tan 0=\dfrac{-1}{2}+\dfrac{C}{1} $
⇒ $ C=\dfrac{1}{2} $
Therefore, $ y(1) $ will be:
⇒ $ y(1)={{\tan }^{-1}}\left[ \dfrac{1}{2}\left( {{1}^{2}}-1 \right)+\dfrac{1}{2{{e}^{{{1}^{2}}}}} \right] $
⇒ $ y(1)={{\tan }^{-1}}\left( \dfrac{1}{2e} \right) $
The correct answer is D. $ {{\tan }^{-1}}\dfrac{1}{2e} $ .

Note: First Order Linear Differential Equation:
A differential equation of the form $ \dfrac{dy}{dx}+P\times y=Q $ , where P and Q are constants or functions of x only, is known as a first order linear differential equation.
Integration by Parts:
 $ \int{f}(x)g(x)dx=\text{ }\!\!~\!\!\text{ }f(x)\text{ }\!\!~\!\!\text{ }\int{\text{ }\!\!~\!\!\text{ }}g(x)dx-\int{\text{ }\!\!~\!\!\text{ }}[{f}'(x)\int{\text{ }\!\!~\!\!\text{ }}g(x)dx]dx $ .
Integration by substitution:
If we substitute $ x=f(t) $ , then $ dx={f}'(t)dt $ and $ \int{f}(x)dx=\text{ }\!\!~\!\!\text{ }\int{f}[f(t)]{f}'(t)dt $ .