Question

Find the solution for the following equation under the initial value of y (pi/2)=1 . y' – 2xy = e^(x^2)cos³ x

Answer 1

To solve. $\mathrm{y'-2xy=e^{x^{2}}cos^{3}x}$

Given. $\mathrm{y\left(\frac{\pi}{2}\right)=1}$

Solution.

Now, $\mathrm{y'-2xy=e^{x^{2}}cos^{3}x}$

$\Rightarrow\mathrm{\frac{dy}{dx}-2xy=e^{x^{2}}cos^{3}x\quad...(1)}$

Then, $\mathrm{I.F.=e^{\int (-2x)dx}=e^{-x^{2}}}$

Multiplying $\mathrm{(1)}$ by $\mathrm{I.F.=e^{-x^{2}}}$, we get:

$\quad \mathrm{\frac{d}{dx}\left(e^{-x^{2}}y\right)=cos^{3}x}$

$\Rightarrow \mathrm{d\left(e^{-x^{2}}y\right)=cos^{3}x\:dx}$

$\Rightarrow \mathrm{d\left(e^{-x^{2}}y\right)=\frac{1}{4}cos3x\:dx+\frac{3}{4}cosx\:dx}$

On integration, we get:

$\quad \mathrm{\int d\left(e^{-x^{2}}y\right)=\frac{1}{4}\int cos3x\:dx+\frac{3}{4}\int cosx\:dx}$

$\Rightarrow \mathrm{e^{-x^{2}}y=\frac{1}{12}sin3x+\frac{3}{4}sinx+C\quad...(2)}$

where $\mathrm{C}$ is constant of integration

When $\mathrm{y\left(\frac{\pi}{2}\right)=1}$, we get:

$\quad \mathrm{e^{-\frac{{\pi}^{2}}{4}}=-\frac{1}{12}+\frac{3}{4}+C}$

$\Rightarrow \mathrm{C=e^{-\frac{{\pi}^{2}}{4}}-\frac{2}{3}}$

Putting $\mathrm{C=e^{-\frac{{\pi}^{2}}{4}}-\frac{2}{3}}$ in $\mathrm{(2)}$:

$\quad \mathrm{e^{-x^{2}}y=\frac{1}{12}sin3x+\frac{3}{4}sinx+e^{-\frac{{\pi}^{2}}{4}}-\frac{2}{3}}$

This is the required solution.