Question #163048

solve the given system of differential equations:

dx/dt=2x+2y

dy/dt=x+3


1
Expert's answer
2021-02-24T07:04:06-0500

solve the given system of differential equations:

{dxdt=2x+2ydydt=x+3\begin{cases} \frac{dx}{dt}=2x+2y \\ \frac{dy}{dt}=x+3 \end{cases}

Solution:

From the second equation:

x=dydt3x=\frac{dy}{dt}-3  (*)

dxdt=d2ydt2\frac{dx}{dt}=\frac{d^2y}{dt^2}

Substitute xx and dxdt\frac{dx}{dt} into the first equation:

d2ydt2=2(dydt3)+2y\frac{d^2y}{dt^2}=2(\frac{dy}{dt}-3 )+2y

d2ydt22dydt2y=6\frac{d^2y}{dt^2}-2\frac{dy}{dt}-2y=-6 (**)

First find the general solution of the corresponding homogeneous equation:

d2y0dt22dy0dt2y0=0\frac{d^2y_0}{dt^2}-2\frac{dy_0}{dt}-2y_0=0

Let's compose and solve the characteristic equation:

λ22λ2=0\lambda^2-2\lambda-2=0

λ1=13\lambda_1=1-\sqrt3 ,  λ2=1+3\lambda_2=1+\sqrt3 .

y0=C1eλ1t+C2eλ2t=C1e(13)t+C2e(1+3)ty_0=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}=C_1e^{(1-\sqrt3)t}+C_2e^{(1+\sqrt3)t} .

Find a typical (specific) solution of the non-homogeneous equation in the form yp=Ay_p=A

Derivatives of the solution: dypdt=0\frac{dy_p}{dt}=0 , d2ypdt2=0\frac{d^2y_p}{dt^2}=0

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

2A=6-2A=-6

A=3A=3

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

y=y0+yp=C1e(13)t+C2e(1+3)t+3y=y_0+y_p=C_1e^{(1-\sqrt3)t}+C_2e^{(1+\sqrt3)t}+3 .

To find xx substitute dydt\frac{dy}{dt} into the (*):

x=dydt3=C1(13)e(13)t+C2(1+3)e(1+3)t3x=\frac{dy}{dt}-3 =C_1(1-\sqrt3)e^{(1-\sqrt3)t}+C_2(1+\sqrt3)e^{(1+\sqrt3)t}-3

Answer: {x=C1(13)e(13)t+C2(1+3)e(1+3)t3y=C1e(13)t+C2e(1+3)t+3\begin{cases} x =C_1(1-\sqrt3)e^{(1-\sqrt3)t}+C_2(1+\sqrt3)e^{(1+\sqrt3)t}-3 \\ y=C_1e^{(1-\sqrt3)t}+C_2e^{(1+\sqrt3)t}+3 \end{cases}



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