Question

 Solve the following system of ODE

(40D+4)y - (2D+1)z = e^-x

(D+8)y - 3z = 5e^-x ; D=d/dx

Answer 1

Solve the following system of ODE

(40D+4)y - (2D+1)z = e^-x

(D+8)y - 3z = 5e^-x ; D=d/dx

$\begin{cases} 40\frac{dy}{dx}+4y-2\frac{dz}{dx}-z=e^{-x} \\ \frac{dy}{dx}+8y-3z=5e^{-x} \end{cases}$

Solution:

From second equation:

$z=\frac13(\frac{dy}{dx}+8y-5e^{-x} )$ (*)

$\frac{dz}{dx}=\frac13(\frac{d^2y}{dx^2}+8\frac{dy}{dx}+5e^{-x})$

Substitute $z$ and $\frac{dz}{dx}$ into the first equation:

$40\frac{dy}{dx}+4y-\frac23(\frac{d^2y}{dx^2}+8\frac{dy}{dx}+5e^{-x})-\frac13(\frac{dy}{dx}+8y-5e^{-x} )=e^{-x}$

$-\frac23\frac{d^2y}{dx^2}+\frac{103}{3}\frac{dy}{dx}+\frac43y=\frac83e^{-x}$

$2\frac{d^2y}{dx^2}-103\frac{dy}{dx}-4y=-8e^{-x}$ (**)

First find the general solution of the corresponding homogeneous equation:

$2\frac{d^2y_0}{dx^2}-103\frac{dy_0}{dx}-4y_0=0$

Let's compose and solve the characteristic equation:

$2\lambda^2-103\lambda-4=0$

$\lambda_1=\frac{103-\sqrt{10641}}{4}$ , $\lambda_2=\frac{103+\sqrt{10641}}{4}$ .

$y_0=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}=C_1e^{\frac{103-\sqrt{10641}}{4}x}+C_2e^{\frac{103+\sqrt{10641}}{4}x}$ .

Find a typical (specific) solution of the non-homogeneous equation in the form $y_p=Ae^{-x}$

Derivatives of the solution: $\frac{dy_p}{dx}=-Ae^{-x}$ , $\frac{d^2y_p}{dx^2}=Ae^{-x}$

Substitute those "solutions" into the (**):

$2Ae^{-x}+103Ae^{-x}-4Ae^{-x}=-8e^{-x}$

$2A+103A-4A=-8$

$A=-\frac{101}{8}$

$y_p=-\frac{101}{8}e^{-x}$ .

Add the typical and the complementary solutions to get the complete solution:

$y=y_0+y_p=C_1e^{\frac{103-\sqrt{10641}}{4}x}+C_2e^{\frac{103+\sqrt{10641}}{4}x}-\frac{101}{8}e^{-x}$ .

To find $z$ substitute $y$ into the (*):

$z=\frac13(C_1\frac{103-\sqrt{10641}}{4}e^{\frac{103-\sqrt{10641}}{4}x}+C_2\frac{103+\sqrt{10641}}{4}e^{\frac{103+\sqrt{10641}}{4}x}+\frac{101}{8}e^{-x}+$

$8(C_1e^{\frac{103-\sqrt{10641}}{4}x}+C_2e^{\frac{103+\sqrt{10641}}{4}x}-\frac{101}{8}e^{-x})-5e^{-x} )=$

$C_1\frac{135-\sqrt{10641}}{12}e^{\frac{103-\sqrt{10641}}{4}x}+C_2\frac{135+\sqrt{10641}}{12}e^{\frac{103+\sqrt{10641}}{4}x}-\frac{249}{8}e^{-x}$

Answer: $\begin{cases} y=C_1e^{\frac{103-\sqrt{10641}}{4}x}+C_2e^{\frac{103+\sqrt{10641}}{4}x}-\frac{101}{8}e^{-x} \\ z=C_1\frac{135-\sqrt{10641}}{12}e^{\frac{103-\sqrt{10641}}{4}x}+C_2\frac{135+\sqrt{10641}}{12}e^{\frac{103+\sqrt{10641}}{4}x}-\frac{249}{8}e^{-x} \end{cases}$