Question

Solve the given initial value problem. y triple prime plus 11 y double prime plus 38 y prime plus 40 y equals 0y′′′+11y′′+38y′+40y=0 y (0 )equals 0y(0)=0​, y prime (0 )equals negative 19y′(0)=−19​, y double prime (0 )equals 117

Answer 1

The linear homogeneous ODE

$y'''+11y''+38y'+40y=0$

has characteristic equation

$r^3+11r^2+38r+40=0$

which factors to

$(r+2)(r+4)(r+5)=0$

and hence has roots at $r=-2,-4,-5$. So the characteristic solution to the ODE is

$y_c=C_1e^{-2x}+C_2e^{-4x}+C_3e^{-5x}$

Use the given initial conditions to solve for each C.

$y(0)=0\implies C_1+C_2+C_3=0$

$y'(0)=-19\implies -2C_1-4C_2-5C_3=-19$

$y''(0)=4C_1+16C_2+25C_3=117$

$\implies C_1=-9,C_2=8,C_3=1$

so that the particular solution is

$\boxed{y(x)=-9e^{-2x}+8e^{-4x}+e^{-5x}}$