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Find the general solution to $y'' - 2y' + 10 = e^x$.

I think that first I need to find the solution to $y'' - 2y' + 10 = 0$, but if I assume $y=e^{rx}$ then I'm left with the equation $r^2 -2r + 10 = 0$ which has no real solutions.

After that I have that the particular solution is $y_p = \frac{1}{9}e^x$ so unless that's wrong all I'm after is tips on how to solve the homogeneous equation.

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2 Answers

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We need to use the method of undetermined coefficients for $y'' - 2y' + 10 = e^x$

First subtract 10 from both sides because we don't have a $y$ in this equation. So our new equation to work with will be $y''-2y'=e^x-10$

Now we are going to find our $Y_h$ first so change the equation to $r^2-2r=0$

Factoring out the $r$, we have $r(r-2)=0$ and when we solve for this equation, we will see that $r=0 $ and $r=2$

So, we have 2 real roots and the $Y_h=C_1e^{0x} +C_2e^{2x}$ which becomes $Y_h=C_1 +C_2e^{2x}$. Now this is only part of the problem, for we have yet to determine the $Y_p$. To make it easier, we need to split the equation.

Therefore, $y''-2y'=e^x-10$ will become

$y''-2y'=e^x$ and $y''-2y'=-10$

Let's start finding the $Y_p$ for $y''-2y'=e^x$

The right hand side of the equation tells me that $Y_p$ is in the form of $Ce^{\lambda}x$ (sorry that x should be with the $\lambda$). The $\lambda$ is the numerical value, and since it's one, then we leave it alone.

So, we find the first and second derivatives of $Y_p =Ce^x$

$Y_p' =Ce^x$

$Y_p'' =Ce^x$

After we obtain the derivatives we must plug it back in $y''-2y'=e^x$

$Ce^x-2Ce^x=e^x$

Simplify and solve for C

$-Ce^x=e^x$

Dividing $e^x$ on both sides, we get

$-C = 1$ and C = -1

As a result, the $Y_p = -e^x$

Now we use the same process for $y''-2y'=-10$, but the only difference is that the right hand side of the equation is in the form of a polynomial. Obviously I can't just use $Y_p = A$ because if I take the derivatives, I will have 0s and that's something I don't want. You do NOT want $0 = -10$. Moreover, I have a 0 degree polynomial in the equation already.

We need a multiplier, so let's have $Y_p = Ax$

The first and second derivatives are

$Y_p' = A$

$Y_p'' = 0$

Now plug in these derivatives back into $y''-2y'=-10$

$-2A = -10$ and by dividing both sides by $-2$, we have $A = 5$. Therefore, $Y_p= 5x$.

Gathering all of my $Y_p$ and $Y_h$ values, the final answer is

$Y_{p+h} = 5x-e^x + C_1 +C_2e^{2x}$

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You have $y''-2y'=e^x-10$, our homogenous equation is $y''-2y'=0$ so our characteristic equation is $r^2-2r=0\Rightarrow r(r-2)=0\Rightarrow r=0,2$, so our homogenous solution is $y_c(x)=A+Be^{2x}$.

So we consider our particular solution, since we have $e^x-10$ on the right hand side we would try $y_p(x)=Ce^x+D$, but the constant appears in our homogenous solution, so we let $y_p(x)=Ce^x+Dx$

Now our whole (general solution) is $y(x)=y_c(x)+y_p(x)=A+Dx+Ce^x+Be^{2x}$

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