Question #161052

Solve the following system of ODE

(48D+4)y - (2D+1)z = e

-x

(D+8)y - 3z = 5e

-x

; D =d/dx


1
Expert's answer
2021-02-23T09:11:18-0500

Solve the following system of ODE

(48D+4)y - (2D+1)z = e

-x

(D+8)y - 3z = 5e

-x

; D =d/dx


{48dydx+4y2dzdxz=exdydx+8y3z=5ex\begin{cases} 48\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=13(dydx+8y5ex)z=\frac13(\frac{dy}{dx}+8y-5e^{-x} ) (*)

dzdx=13(d2ydx2+8dydx+5ex)\frac{dz}{dx}=\frac13(\frac{d^2y}{dx^2}+8\frac{dy}{dx}+5e^{-x})

Substitute zz and dzdx\frac{dz}{dx} into the first equation:

48dydx+4y23(d2ydx2+8dydx+5ex)13(dydx+8y5ex)=ex48\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}

23d2ydx2+1273dydx+43y=83ex-\frac23\frac{d^2y}{dx^2}+\frac{127}{3}\frac{dy}{dx}+\frac43y=\frac83e^{-x}

2d2ydx2127dydx4y=8ex2\frac{d^2y}{dx^2}-127\frac{dy}{dx}-4y=-8e^{-x} (**)

First find the general solution of the corresponding homogeneous equation:

2d2y0dx2127dy0dx4y0=02\frac{d^2y_0}{dx^2}-127\frac{dy_0}{dx}-4y_0=0

Let's compose and solve the characteristic equation:

2λ2127λ4=02\lambda^2-127\lambda-4=0

λ1=127161614\lambda_1=\frac{127-\sqrt{16161}}{4} , λ2=127+161614\lambda_2=\frac{127+\sqrt{16161}}{4} .

y0=C1eλ1x+C2eλ2x=C1e127161614x+C2e127+161614xy_0=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}=C_1e^{\frac{127-\sqrt{16161}}{4}x}+C_2e^{\frac{127+\sqrt{16161}}{4}x} .

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

Derivatives of the solution: dypdx=Aex\frac{dy_p}{dx}=-Ae^{-x} , d2ypdx2=Aex\frac{d^2y_p}{dx^2}=Ae^{-x}

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

2Aex+127Aex4Aex=8ex2Ae^{-x}+127Ae^{-x}-4Ae^{-x}=-8e^{-x}

2A+127A4A=82A+127A-4A=-8

A=8125A=-\frac{8}{125}

yp=8125exy_p=-\frac{8}{125}e^{-x} .

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

y=y0+yp=C1e127161614x+C2e127+161614x8125exy=y_0+y_p=C_1e^{\frac{127-\sqrt{16161}}{4}x}+C_2e^{\frac{127+\sqrt{16161}}{4}x}-\frac{8}{125}e^{-x} .

To find zz substitute yy into the (*):

z=13(C1127161614e127161614x+C2127+161614e127+161614x+8125ex+z=\frac13(C_1\frac{127-\sqrt{16161}}{4}e^{\frac{127-\sqrt{16161}}{4}x}+C_2\frac{127+\sqrt{16161}}{4}e^{\frac{127+\sqrt{16161}}{4}x}+\frac{8}{125}e^{-x}+

8(C1e127161614x+C2e127+161614x8125ex)5ex)=8(C_1e^{\frac{127-\sqrt{16161}}{4}x}+C_2e^{\frac{127+\sqrt{16161}}{4}x}-\frac{8}{125}e^{-x})-5e^{-x} )=

C11591616112e127161614x+C2159+1616112e127+161614x227125exC_1\frac{159-\sqrt{16161}}{12}e^{\frac{127-\sqrt{16161}}{4}x}+C_2\frac{159+\sqrt{16161}}{12}e^{\frac{127+\sqrt{16161}}{4}x}-\frac{227}{125}e^{-x}

Answer: {y=C1e127161614x+C2e127+161614x8125exz=C11591616112e127161614x+C2159+1616112e127+161614x227125ex\begin{cases} y=C_1e^{\frac{127-\sqrt{16161}}{4}x}+C_2e^{\frac{127+\sqrt{16161}}{4}x}-\frac{8}{125}e^{-x} \\ z=C_1\frac{159-\sqrt{16161}}{12}e^{\frac{127-\sqrt{16161}}{4}x}+C_2\frac{159+\sqrt{16161}}{12}e^{\frac{127+\sqrt{16161}}{4}x}-\frac{227}{125}e^{-x} \end{cases}

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