- Thread starter
- Admin
- #1

Here is my working:

We are given the ODE:

(1) $\displaystyle y''+3y'-10y=xe^x+x+1$

Since $\displaystyle (D-1)^2$ annihilates $\displaystyle xe^x$ and $\displaystyle D^2$ annihilates $\displaystyle x+1$ then:

$\displaystyle A\equiv (D(D-1))^2$

annihilates $\displaystyle xe^x+x+1$.

Therefore, applying $\displaystyle A$ to both sides of (1) yields:

$\displaystyle A[y''+3y'-10y]=A[xe^x+x+1]$

$\displaystyle (D(D-1))^2(D^2+3D-10)[y]=0$

(2) $\displaystyle (D(D-1))^2(D+5)(D-2)[y]=0$

The auxiliary equation associated with (2) is:

$\displaystyle r^2(r-1)^2(r+5)(r-2)=0$

which has the roots:

$\displaystyle r=-5,\,0,\,1,\,2$ where the roots $\displaystyle r=0,\,1$ are repeated roots, i.e., of multiplicity 2.

Hence, a general solution to (2) is:

(3) $\displaystyle y(x)=c_1+c_2x+c_3e^x+c_4xe^x+c_5e^{2x}+c_6e^{-5x}$

Now, recall that a general solution to (1) is of the form $\displaystyle y(x)=y_h(x)+y_p(x)$. Since every solution to (1) is also a solution to (2), then $\displaystyle y(x)$ must have the form displayed on the right-hand side of (3). But, we recognize that:

$\displaystyle y_h(x)=c_5e^{2x}+c_6e^{-5x}$

and so there must exist a particular solution of the form:

$\displaystyle y_p(x)=c_1+c_2x+c_3e^x+c_4xe^x$

In order to substitute this into (1), we must first compute:

$\displaystyle y_p'(x)=c_2+c_3e^x+c_4e^x(x+1)$

$\displaystyle y_p''(x)=c_3e^x+c_4e^x(x+2)$

and so we find:

$\displaystyle (c_3e^x+c_4e^x(x+2))+3(c_2+c_3e^x+c_4e^x(x+1))-10(c_1+c_2x+c_3e^x+c_4xe^x)=xe^x+x+1$

Collecting like terms, we may write:

$\displaystyle (-6c_4)xe^x+(5c_4-6c_3)e^x+(-10c_2)x+(3c_2-10c_1)=(1)xe^x+(0)e^x+(1)x+(1)$

Equating coefficients yields:

$\displaystyle -6c_4=1\,\therefore\,c_4=-\frac{1}{6}$

$\displaystyle 5c_4-7c_3=0\,\therefore\,c_3=-\frac{5}{36}$

$\displaystyle -10c_2=1\,\therefore\,c_2=-\frac{1}{10}$

$\displaystyle 3c_2-10c_1=1\,\therefore\,c_1=-\frac{13}{100}$

Thus, we have:

$\displaystyle y_p(x)=-\frac{13}{100}-\frac{1}{10}x-\frac{5}{36}e^x-\frac{1}{6}xe^x$

and so, the general solution to (1) is:

$\displaystyle y(x)=c_1e^{2x}+c_2e^{-5x}-\frac{13}{100}-\frac{1}{10}x-\frac{5}{36}e^x-\frac{1}{6}xe^x$