solve the given system of differential equations:
{dtdx=2x+2ydtdy=x+3
Solution:
From the second equation:
x=dtdy−3 (*)
dtdx=dt2d2y
Substitute x and dtdx into the first equation:
dt2d2y=2(dtdy−3)+2y
dt2d2y−2dtdy−2y=−6 (**)
First find the general solution of the corresponding homogeneous equation:
dt2d2y0−2dtdy0−2y0=0
Let's compose and solve the characteristic equation:
λ2−2λ−2=0
λ1=1−3 , λ2=1+3 .
y0=C1eλ1t+C2eλ2t=C1e(1−3)t+C2e(1+3)t .
Find a typical (specific) solution of the non-homogeneous equation in the form yp=A
Derivatives of the solution: dtdyp=0 , dt2d2yp=0
Substitute those "solutions" into the (**):
−2A=−6
A=3
Add the typical and the complementary solutions to get the complete solution:
y=y0+yp=C1e(1−3)t+C2e(1+3)t+3 .
To find x substitute dtdy into the (*):
x=dtdy−3=C1(1−3)e(1−3)t+C2(1+3)e(1+3)t−3
Answer: {x=C1(1−3)e(1−3)t+C2(1+3)e(1+3)t−3y=C1e(1−3)t+C2e(1+3)t+3
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