Question

y¨ + 4y´ + 4y=(3 + x)e^-2x , y(0)=2, y´(0)=5

Answer 1

Let us solve the differential equation $y'' + 4y' + 4y=(3 + x)e^{-2x} ,\ y(0)=2, y'(0)=5$.

The characteristic equation $k^2+4k+4=0$ is equivalent to $(k+2)^2=0,$ and hence has the solutions $k_1=k_2=-2.$ Therefore, the general solution of the differential equation $y'' + 4y' + 4y=(3 + x)e^{-2x}$ is $y=(C_1+C_2x)e^{-2x}+y_p,$ where $y_p=x^2(ax+b)e^{-2x}=(ax^3+bx^2)e^{-2x}.$

It follows that

$y_p'=(3ax^2+2bx)e^{-2x}-2(ax^3+bx^2)e^{-2x}\\=(3ax^2+2bx-2ax^3-2bx^2)e^{-2x},$

$y''=(6ax+2b-6ax^2-4bx)e^{-2x}-2(3ax^2+2bx-2ax^3-2bx^2)e^{-2x}\\= (6ax+2b-6ax^2-4bx-6ax^2-4bx+4ax^3+4bx^2)e^{-2x}.$

Then we have

$(6ax+2b-6ax^2-4bx-6ax^2-4bx+4ax^3+4bx^2)e^{-2x}+4(3ax^2+2bx-2ax^3-2bx^2)e^{-2x}+4(ax^3+bx^2)e^{-2x}= (3 + x)e^{-2x},$

and hence $6ax+2b=3 + x.$ We conclude that $6a=1,\ 2b=3,$ and thus $a=\frac{1}6,\ b=\frac{3}2.$

Consequently, the general solution is of the form

$y=(C_1+C_2x)e^{-2x}+(\frac{1}6x^3+\frac{3}2x^2)e^{-2x}=(C_1+C_2x+\frac{3}2x^2+\frac{1}6x^3)e^{-2x}.$

Then $2=y(0)=C_1.$

It follows that $y'=(C_2+3x+\frac{1}2x^2)e^{-2x}-2(C_1+C_2x+\frac{3}2x^2+\frac{1}6x^3)e^{-2x},$ and hence $5=y'(0)=C_2-2C_1=C_2-4.$ Then $C_2=9.$

We conclude that the solution of the differential equation $y'' + 4y' + 4y=(3 + x)e^{-2x} ,\ y(0)=2, y'(0)=5$ is

$y=(2+9x+\frac{3}2x^2+\frac{1}6x^3)e^{-2x}.$