Differential equation exercise asked in the Csb.gov.lb (مباراة تعيين استاذ تعليم ثانوي 2004 ) the last exercise

Exercise :

1) Determine the real numbers $a,b\,\And c$ such that

                $\int{{{x}^{2}}{{e}^{-x}}dx=\left( a{{x}^{2}}+bx+c \right){{e}^{-x}}+k}$  where $k\in \mathbb{R}$

2) Consider the differential equation

       $\left( E \right):\,\,\,\,{{x}^{2}}\frac{dy}{dx}-\left( {{x}^{2}}-4x \right)y+1=0$

Take $z={{x}^{4}}{{e}^{-x}}y$ then find the differential equation $\left( E' \right)$ satisfied by $z$ and

Deduce the General solution of $\left( E \right)$

Solution:

1) To find the real numbers $a,b\,\And \,c$ we need to integrate $\int{{{x}^{2}}{{e}^{-x}}dx}$

Let $u={{x}^{2}}\,\,\,\And \,\,\,dv={{e}^{-x}}dx\Leftrightarrow \,du=2x\,dx\,\,\And \,\,v=-{{e}^{-x}}$

So $\int{{{x}^{2}}{{e}^{-x}}dx}=-{{x}^{2}}{{e}^{-x}}+2\int{x{{e}^{-x}}dx}$

Also let $U=x\,\,\And \,\,dV={{e}^{-x}}\Leftrightarrow dU=dx\,\,\And \,\,V=-{{e}^{-x}}$

So $\int{x{{e}^{-x}}dx=-x{{e}^{-x}}+\int{{{e}^{-x}}dx=-x{{e}^{-x}}-{{e}^{-x}}+k}}$

Hence $\int{{{x}^{2}}{{e}^{-x}}dx=-{{x}^{2}}{{e}^{-x}}-2x{{e}^{-x}}-2{{e}^{-x}}+k}={{e}^{-x}}\left( -{{x}^{2}}-2x-2 \right)+k$

Thus $a=-1\,\,,\,b=-2\,\,\And \,\,c=-2$

2) We have $z={{x}^{4}}{{e}^{-x}}y\Leftrightarrow y=\frac{z}{{{x}^{4}}{{e}^{-x}}}=z{{x}^{-4}}{{e}^{x}}$

$\frac{dy}{dx}=\frac{d}{dx}\left( z{{x}^{-4}}{{e}^{x}} \right)=\frac{d}{dx}\left( z{{x}^{-4}} \right){{e}^{x}}+\frac{d}{dx}\left( {{e}^{x}} \right)z{{x}^{-4}}$

     $=\left( \frac{d}{dx}\left( z \right){{x}^{-4}}+\frac{d}{dx}\left( {{x}^{-4}} \right)z \right){{e}^{x}}+z{{x}^{-4}}{{e}^{x}}={{e}^{x}}\left( \frac{dz}{dx}{{x}^{-4}}-4{{x}^{-5}}z+z{{x}^{-4}} \right)$

Substitute $\frac{dy}{dx}\,\,\And \,y$ in the differential equation $\left( E \right)$ to obtain$\left( E' \right)$

${{x}^{2}}\left( z'{{x}^{-4}}-4{{x}^{-5}}z+{{x}^{-4}}z \right){{e}^{x}}-\left( {{x}^{2}}-4x \right)\left( z{{x}^{-4}}{{e}^{x}} \right)+1=0$

${{e}^{x}}\left( z'{{x}^{-2}}-4{{x}^{-3}}z+{{x}^{-2}}z-{{x}^{-2}}z+4{{x}^{-3}}z \right)+1=0$

$\Leftrightarrow z'{{x}^{-2}}{{e}^{x}}+1=0\Leftrightarrow \frac{dz}{dx}=\frac{-1}{{{x}^{-2}}{{e}^{x}}}=-{{x}^{2}}{{e}^{-x}}\Leftrightarrow dz=-{{x}^{2}}{{e}^{-x}}dx$

So $z=-\int{{{x}^{2}}{{e}^{-x}}dx}=-{{e}^{-x}}\left( -{{x}^{2}}-2x-2 \right)+k$  is the General solution of $\left( E' \right)$

But $z={{x}^{4}}y{{e}^{-x}}\Leftrightarrow {{e}^{-x}}{{x}^{4}}y=-{{e}^{-x}}\left( -{{x}^{2}}-2x-2 \right)+k$  multiply by ${{e}^{x}}$

$\Leftrightarrow {{x}^{4}}y={{x}^{2}}+2x+2+k{{e}^{x}}\Leftrightarrow y\left( x \right)=\frac{{{x}^{2}}+2x+2}{{{x}^{4}}}+\frac{k{{e}^{x}}}{{{x}^{4}}}$ is the General solution of $\left( E \right)$