Answer to Question #163211 in Differential Equations for Divine Grace Jaravata

Question

.  y”  -  y’ -   2y  =  1 – 2x - 9e^-x        

Answer 1

Solution:

We know that the general solution will be the sum of the complementary solution and particular solution.

First we find the complementary solution by solving $\frac{d^{2} y(x)}{d x^{2}}-\frac{d y(x)}{d x}-2 y(x)=0$ :

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$ .

Substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)-\frac{d}{d x}\left(e^{\lambda x}\right)-2 e^{\lambda x}=0$

$\lambda^{2} e^{\lambda x}-\lambda e^{\lambda x}-2 e^{\lambda x}=0$

$\left(\lambda^{2}-\lambda-2\right) e^{\lambda x}=0$

Since $e^{\lambda x} \neq 0$ for any finite $\lambda$ , the zeros must come from the polynomial:

$\lambda^{2}-\lambda-2=0$

$(\lambda-2)(\lambda+1)=0$

$\lambda=-1 \ or\ \lambda=2$

The general solution is the sum of the above solutions:

$y(x)=y_{1}(x)+y_{2}(x)=c_{1} e^{-x}+c_{2} e^{2 x}$

Now we determine the particular solution by the method of undetermined coefficients:

The particular solution will be the sum of the particular solutions to $\frac{d^{2} y(x)}{d x^{2}}-\frac{d y(x)}{d x}-2 y(x)=-2 x+1\ and\ \frac{d^{2} y(x)}{d x^{2}}-\frac{d y(x)}{d x}-2 y(x)=-9 e^{-x}$

The particular solution to $\frac{d^{2} y(x)}{d x^{2}}-\frac{d y(x)}{d x}-2 y(x)=-2 x+1$ is of the form:

$y_{p_{1}}(x)=a_{1}+a_{2} x$

The particular solution to $\frac{d^{2} y(x)}{d x^{2}}-\frac{d y(x)}{d x}-2 y(x)=-9 e^{-x}$ is of the form:

$y_{p_{2}}(x)=x\left(a_{3} e^{-x}\right)$ , where $a_{3} e^{-x}$ was multiplied by $x$ to account for $e^{-x}$ in the complementary solution.

$y_{p}(x)=y_{p_{1}}(x)+y_{p_{2}}(x)=a_{1}+a_{2} x+a_{3} e^{-x} x$

Now, we solve for the unknown constants $a_{1}, a_{2},\ and\ a_{3}$ :

$\begin{array}{l} \text { Compute } \frac{d y_{p}(x)}{d x}: \\ \frac{d y_{p}(x)}{d x}=\frac{d}{d x}\left(a_{1}+a_{2} x+a_{3} e^{-x} x\right) \\ =a_{2}+a_{3} e^{-x}-a_{3} e^{-x} x \\ \text { Compute } \frac{d^{2} y_{p}(x)}{d x^{2}}: \\ \frac{d^{2} y_{p}(x)}{d x^{2}}=\frac{d^{2}}{d x^{2}}\left(a_{1}+a_{2} x+a_{3} e^{-x} x\right) \\ =a_{3}\left(-2 e^{-x}+e^{-x} x\right) \end{array}$

Substitute the particular solution $y_{p}(x)$ into the differential equation:

$\frac{d^{2} y_{p}(x)}{d x^{2}}-\frac{d y_{p}(x)}{d x}-2 y_{p}(x)=-9 e^{-x}-2 x+1$

$\begin{array}{l} a_{3}\left(-2 e^{-x}+e^{-x} x\right)-\left(a_{2}+a_{3} e^{-x}-a_{3} e^{-x} x\right)-2\left(a_{1}+a_{2} x+a_{3} e^{-x} x\right) \\ \quad=-9 e^{-x}-2 x+1 \end{array}$

$-2 a_{1}-a_{2}-3 a_{3} e^{-x}-2 a_{2} x=1-9 e^{-x}-2 x$

On equating the coefficients on both sides of the equation:

$-2 a_{1}-a_{2}=1$

$-3 a_{3}=-9$

$-2 a_{2}=-2$

On solving,

$\begin{array}{l} a_{1}=-1 \\ a_{2}=1 \\ a_{3}=3 \end{array}$

Substitute

$a_{1}, a_{2},\ and\ a_{3}\ into\ y_{p}(x)=a_{1}+x a_{2}+e^{-x} x a_{3} \\ We\ get\ \\ y_{p}(x)=x+3 e^{-x} x-1$

So, the general solution is: $y(x)=y_{c}(x)+y_{p}(x)=x+3 e^{-x} x+c_{1} e^{-x}+c_{2} e^{2 x}-1$