# 2nd order inhomogeneous DE, particular solution

$$2y''-5y'-3y=2e^{3x} .$$ When I was doing the particular solution, I got $$y_p=Ae^{3x} , \, y'_P=3Ae^{3x} , \, y''_P = 9Ae^{3x} .$$ Substituting them into the DE, I get $18Ae^{3x}-15Ae^{3x}-3Ae^{3x}=2e^{3x}$. I tried to work out $A$, but the LHS becomes 0, and I can't work out the particular solution. Does it means this DE has no particular solution?

• $Ae^{3x}$ is in fact a solution to the homogeneous equation, which means it can't be a particular solution of the inhomogenous equation. This means you have to get more creative inventive trial functions to test. Commented Mar 15, 2014 at 9:30
• $B e^{-x/2}$ is another one Commented Mar 15, 2014 at 9:36
• if you end up with a $0 = 2e^{3x}$, that means that your $y_p$ doesn't work and you have to use a multiplier. Just multiply your $y_p$ by $x$ and repeat the process again until you can solve for $A$. Commented Mar 15, 2014 at 11:39

Try $y(x) = Axe^{3x}$. Then, $y'(x) = Ae^{3x} + 3Axe^{3x}$ and $y''(x) = 3Ae^{3x} + 3Ae^{3x} + 9Axe^{3x}$, so that your ODE yields $$2 \cdot \left( 6Ae^{3x} + 9Axe^{3x} \right) - 5 \cdot \left( Ae^{3x} + 3Axe^{3x} \right) -3 \cdot Axe^{3x} = 2e^{3x} .$$ Thus, you arrive at the simultaneous equations $12A - 5A = 2$ and $(18A - 15A - 3A)x = 0x$. The first gives you that $A = \frac{2}{7}$, and the second is a trivial identity that is satisfied for any $A$. Hence, a particular solution is $y_p(x) = \frac{2}{7} xe^{3x}$.

Usually, the particular solution to this equation is $Ce^{3x}$. But, this solution is also a solution of the homogenous one i.e. we have a duplication with the homogeneous solution. In this case, we multiple $Ce^{3x}$ by $x$ until we have no duplication with the homogenous solution. So, the two solution will be independent. Thus your particular solution should be $Cxe^{3x}$.

Moreover,(Not your case) if the new suggested particular solution occurs in the homogenous equation, multiply the suggested solution again by $x$. For example consider the equation $$y''+8y'+16=e^{-4x}$$

• Oh I see, thanks for telling me that Commented Mar 16, 2014 at 19:40

The complex Laplace transform can be useful to find the general solution. Thus let $x\geqslant 0$ (for $x<0$ we can proceed similarly) and

$$\hat{y}(z)=\int_{0}^{\infty }dx\exp [izx]y(x),\;{\text{Im}}z>0.$$ Then, with $z=\omega +i\delta$, $\delta >0$, and $\theta (x)$ the Heaviside step function,

$$\hat{y}(\omega +i\delta )=\int_{0}^{\infty }dx\exp [i(\omega +i\delta )x]y(x)=\int_{-\infty }^{\infty }dx\exp [i\omega x]\theta (x)\exp [-\delta x]y(x),$$ so $$\begin{eqnarray*} \theta (x)\exp [-\delta x]y(x) &=&\frac{1}{2\pi }\int_{-\infty }^{\infty }d\omega \exp [-i\omega x]\hat{y}(\omega +i\delta ) \\ y(x) &=&\frac{1}{2\pi }\exp [\delta x]\int_{-\infty }^{\infty }d\omega \exp [-i\omega x]\hat{y}(\omega +i\delta )=\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izx]\hat{y}(z), \end{eqnarray*}$$ where $\Gamma =\mathbb{R}+i\delta$. Since $$\begin{eqnarray*} \int_{0}^{\infty }dx\exp [izx]\partial _{x}^{2}y(x) &=&-(\partial _{x}y)(0)+izy(0)-z^{2}\hat{y}(z), \\ \int_{0}^{\infty }dx\exp [izx]\partial _{x}y(x) &=&-y(0)-iz\hat{y}(z), \\ \int_{0}^{\infty }dx\exp [izx]\exp [3x] &=&-\frac{1}{iz+3}, \end{eqnarray*}$$ we obtain

$$\hat{y}(z)=\frac{1}{z^{2}-5iz-15}\{\frac{1}{iz+3}-2(\partial _{x}y)(0)-(5-2iz)y(0)\}$$ from which $y(x)$ can be obtained. We see that the general solution depends on $(\partial _{x}y)(0)$ and $y(0)$.

Note: There may be mistakes in my calculations (as is usually the case) but the general idea will be clear.