Question

The solution of the differential equation $\frac{d^{2}y}{d{x}^2}+3y=-2x$ is.

• A. $c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x-\frac{2}{3}x$
• B. $c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x-\frac{4}{5}$
• C. $c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x-2x^2+\frac{4}{9}$
• D. $c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x-\frac{2}{3}{x}^2+\frac{4}{9}$

Answer 1

The correct option is A $c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x-\frac{2}{3}x$

$\frac{d^{2}y}{d{x}^2}+3y=-2x.......\left(i\right)$

This is second order non homogeneous differential equation.

Its solution is given as $CF+PI$

For $CF$ part

$\frac{d^{2}y}{d{x}^2}+3y=\mathrm{0....}\left(ii\right)y={c_{1}e}^{mx}\frac{dy}{dx}=c_{1}m{e}^{mx}\frac{d^{2}y}{d{x}^2}=c_1{m}^2{e}^{mx}$

Substituting in $\left(ii\right)$, we get
$m^2{e}^{mx}+3e^{mx}=0e^{mx}(m^2+3)=0m^2+3=0\Rightarrow m=\sqrt{-3}m=i\sqrt{3}y=c_1{e}^{i\sqrt{3}x}=c_1(\cos \sqrt{3}x+i\sin \sqrt{3}x)y=c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x$

For $PI$ part

Let $y=cx+d$

$\frac{dy}{dx}=c\frac{d^{2}y}{d{x}^2}=0$

Substituting in $\left(i\right)$

$0+3(cx+d)=-2x3cx+3d=-2x$

Comparing both sides

$c=-\frac{2}{3},d=0$

So solution for $PI$ part is

$y=-\frac{2}{3}x$

General solution is $CF+PI$

$c_1\cos \sqrt{3}x+c_2\sin \sqrt{3}x-\frac{2}{3}x$