Question

Verify that the given functions form the fundamental set of solution of the differential equation on the given indicated interval.

y" — 7y' + 10y = 24e^x
y = Cl1e^2x + C2e^5x + 6e^x, (-0, infinity)

Answer 1

Solution;

The symbolic form of the equation is;

$(D^2-7D+10)y=24e^x$

The auxiliary equation is;

$m^2-7m+10=0$

Rewrite as the following;

$m^2-2m-5m+10=0$

$m(m-2)-5(m-2)=0$

$(m-2)(m-5)=0$

m=2 or 5

Hence;

$C.F=y_c$ =$C_1e^{2x}+C_2e^{5x}$

To find the particular solution;

Take,

$y_1=e^{2x}$ ,$y_2=e^{5x}$ and $X=24e^x$

Find the Workisian;

$W=\begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}$ =$\begin{bmatrix} e^{2x} & e^{5x} \\ 2e^{2x} & 5e^{5x} \end{bmatrix}$

$W=5e^{7x}-2e^{7x}=3e^{7x}$

$P.I=y_p=uy_1+vy_2$

Take ;

$u=-\int\frac{y_2X}{W}=-\int\frac{e^{5x}24e^x}{3e^{7x}}=8e^{-x}$

$v=\int\frac{y_1X}{W}=\int\frac{e^{2x}24e^x}{3e^{7x}}=-2e^{-4x}$

$P.I=y_p=8e^{-x}e^{2x}+-2e^{-4x}e^{5x}$

$y_p=8e^x-2e^x=6e^x$

The general solution is;

$y=C.F+P.I=y_c+y_p$

$y=C_1e^{2x}+C_2e^{5x}+6e^x$

Hence the solution is verified.