Answer to Question #294113 in Differential Equations for Yohana Andrew

Question

Solve the initial value problem

y"-3y'-10y=0, y'(0)=1, y'(0)=2

Answer 1

"y"-3y'-10y=0, y(0)=1, y'(0)=2"

Auxiliary equation is

$k^2-3k-10=0$

$k=\frac{3\pm \sqrt{9+40}}{2}$

$k_1=-2,k_2=5$

General solution is

$y=c_1e^{-2x}+c_2e^{5x}...(1)$

Since, $y(0)=1$

$1=c_1+c_2...(2)$

And

$y'=-2c_1e^{-2x}+5c_2e^{5x}$

Also $y'(0)=2$

$2=-2c_1+5c_2...(3)$

Multiply equation (2) with 2 and add with equation (3), we get

$4=7c_2\Rightarrow c_2=\frac{4}{7}$

From equation (2),

$1=c_1+\frac 4 7\Rightarrow c_1=\frac 3 7$

$y=\frac 3 7e^{-2x}+\frac 4 7e^{5x}\Rightarrow 7y=3e^{-2x}+4e^{5x}$