solving the differential equation $(y''- 2y' - 3y = 2xe^{2x})$

Question

ok so i have this equation $y''- 2y' - 3y = 2xe^{2x}$. I have tried everything to find the right function to work with for example : i tried $Y(t) = Ax^3e^{2x}$ but it doesn't solve my equation tried with every power possible on $t$ . Am i missing a methodology or something ? can someone help me with this please ?

Answer 1

hint

To eliminate the exponential term, Put $$y=ze^{2x}$$

then

$$y'=(z'+2z)e^{2x}$$ $$y''=(z''+4z'+4z)e^{2x}$$

the equation becomes

$$z''+2z'-3z=2x$$

without RHS, the solution is $$z_h=Ae^x+Be^{-3x}$$

and the Polynomial particular solution $$z_p=-\frac 23x-\frac 49$$

The general solution is $$y_G=(z_h+z_p)e^{2x}$$

Answer 2

Hint:

You have to

• find the general solution of the associated homogeneous equation $\;y''-2y'-3y=0$.
• Find a particular solution a of the complete (non-homogeneous) equation, which will have the form \begin{cases} (ax+b)\mathrm e^{2x}&\text{if $2$ is not a root of the characteristic equation,} \\ x(ax+b)\mathrm e^{2x}& \text{if $2$ is a simple root of the characteristic equation,} \\ x^2(ax+b)\mathrm e^{2x}& \text{if $2$ is a double root of the characteristic equation.} \end{cases}
• Add this particular solution to the general solution of the homogeneous equation.

Answer 3

Here's a solution following the standard algorithm that Bernard suggests in his answer.

Solve the homogenous equation $$y'' - 2y' - 3y = 0$$ to find $$y_H(x) = c_1e^{-x} + c_2e^{3x}.$$

Since $e^{2x}$ does not appear in the general solution, try $Y = (Ax+B)e^{2x}$, since it mimicks the form of the non-homogeneous term. Its derivatives are

$$Y' = Ae^{2x} + 2Y$$

$$Y'' = 2Ae^{2x} + 2Y'$$

Substitute into the differential equation to find

$$\left(2Ae^{2x} + 2Y'\right) - 2\left(Ae^{2x} + 2Y\right) -3Y = 2xe^{2x}$$ $$2Y' - 7Y = 2xe^{2x}$$ $$2Ae^{2x} - 3Y = 2xe^{2x}$$ $$2A - 3(Ax+B) = 2x$$ $$(2+3A)x + (3B-2A) = 0$$

If this last equation is to hold for every $x$, then $2+3A=0$ and $3B-2A=0$. Thus $A=-2/3$ and $B = 2A/3 = -4/9$, so

$$Y(x) = -\left(\frac{2}{3}x + \frac{4}{9}\right)e^{2x}$$