Second-order nonhomogeneous differential equations initial value problems

Steps for solving a second-order nonhomogeneous differential equation initial value problem

To solve an initial value problem for a second-order nonhomogeneous differential equation, we’ll follow a very specific set of steps.

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We can summarize these steps as follows:

1. Find the complementary solution $y_c(x)$.

2. Find the particular solution $y_p(x)$.

3. Put them together to find the general solution $y(x)=y_c(x)+y_p(x)$.

4. Find the derivative of the general solution $y'(x)=y_c'(x)+y_p'(x)$.

5. Plug the given initial conditions into the general solution and its derivative to create a system of linear equations.

6. Solve the system in order to find values for $c_1$ and $c_2$.

7. Plug $c_1$ and $c_2$ back into the general solution to solve the initial value problem.

Example of a nonhomogeneous differential equation with an exponential function

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Finding the general solution of the second-order nonhomogeneous differential equation given a set of initial conditions

Example

Find the general solution of the differential equation, if $y(0)=0$ and $y'(0)=1$.

$y''+y'=x-2e^{-x}$

First we need to work on the complementary solution, which we’ll do by substituting $0$ for the entire right side and focusing only the left side.

$y''+y'=0$

Then we’ll make the substitution $y'=r$.

$r^2+r=0$

$r(r+1)=0$

So the roots are

$r_1=0$

$r_2=-1$

These are distinct real roots, so we’ll use the formula for the complementary solution with distinct real roots and get

$y_c(x)=c_1e^{r_1x}+c_2e^{r_2x}$

$y_c(x)=c_1e^{(0)x}+c_2e^{-1x}$

$y_c(x)=c_1(1)+c_2e^{-x}$

$y_c(x)=c_1+c_2e^{-x}$

We’ll hold on to the complementary solution and switch over to the particular solution. The first thing we notice is that we have a polynomial function, $x$, and an exponential function, $-2e^{-x}$. We’ll use $Ax+B$ as our guess for the polynomial function, and we’ll use $Ce^{-x}$ as our guess for the exponential function. Putting these together, our guess for the particular solution will be

$y_p(x)=Ax+B+Ce^{-x}$

Comparing this to the complementary solution, we can see that $c_2e^{-x}$ from the complementary solution and $Ce^{-x}$ from the particular solution are overlapping terms. To fix this, we’ll multiply $Ce^{-x}$ from the particular solution by $x$, such that our guess becomes

$y_p(x)=Ax+B+Cxe^{-x}$

We also need to realize that $c_1$ from the complementary solution and $B$ from the particular solution are overlapping terms. To fix this, we’ll multiply the polynomial portion of the particular solution by $x$, such that our guess becomes

$y_p(x)=Ax^2+Bx+Cxe^{-x}$

Taking the first and second derivatives of this guess, we get

$y_p'(x)=2Ax+B+Ce^{-x}-Cxe^{-x}$

$y_p''(x)=2A-2Ce^{-x}+Cxe^{-x}$

Plugging the first two derivatives into the original differential equation, we get

$2A-2Ce^{-x}+Cxe^{-x}+2Ax+B+Ce^{-x}-Cxe^{-x}=x-2e^{-x}$

$(2A+B)+2Ax-Ce^{-x}=x-2e^{-x}$

To solve an initial value problem for a second-order nonhomogeneous differential equation, we’ll follow a very specific set of steps.

Equating coefficients from the left and right side, we get

$2A+B=0$

$2\left(\frac12\right)+B=0$

$B=-1$

and

$2A=1$

$A=\frac12$

and

$-C=-2$

$C=2$

We’ll plug the results into our guess for the particular solution to get

$y_p(x)=\frac12x^2-x+2xe^{-x}$

Putting this together with the complementary solution gives us the general solution to the differential equation.

$Y(x)=y_c(x)+y_p(x)$

$Y(x)=c_1+c_2e^{-x}+\frac12x^2-x+2xe^{-x}$

Now we’ll take the derivative of the general solution.

$Y'(x)=-c_2e^{-x}+x-1+2e^{-x}-2xe^{-x}$

We can now plug the given initial conditions $y(0)=0$ and $y'(0)=1$ into the general solution and its derivative to create a system of linear equations.

$Y(x)=c_1+c_2e^{-x}+\frac12x^2-x+2xe^{-x}$

$0=c_1+c_2e^{-(0)}+\frac12(0)^2-0+2(0)e^{-(0)}$

$0=c_1+c_2(1)$

$0=c_1+c_2$

and

$Y'(x)=-c_2e^{-x}+x-1+2e^{-x}-2xe^{-x}$

$1=-c_2e^{-(0)}+0-1+2e^{-(0)}-2(0)e^{-(0)}$

$1=-c_2(1)-1+2(1)$

$1=-c_2+1$

$0=-c_2$

$0=c_2$

Plugging $c_2=0$ into $0=c_1+c_2$ to solve the system of linear equations gives

$0=c_1+0$

$0=c_1$

Plugging these values for $c_1$ and $c_2$ back into the general solution gives

$Y(x)=0+(0)e^{-x}+\frac12x^2-x+2xe^{-x}$

$Y(x)=\frac12x^2-x+2xe^{-x}$

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