Ytukyg

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 ?

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
$$y_G=(z_h+z_p)e^{2x}$$

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.

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

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

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

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