Question

Solve the following IVP for the second order linear equations y''-4y'+9y=0, y(0)=0, y'(0)=-8

Second order linear equations
ANSWERED
asked 2021-01-10
Solve the following IVP for the second order linear equations
\(y''-4y'+9y=0,\ \ y(0)=0,\ \ y'(0)=-8\)

Answers (1)

2021-01-11

Given:
\(y''-4y'+9y=0\)
\(y(0)=0\)
\(y'(0)=-8\)
The auxiliary equation of the givendifferential equation is
\(m^2-4m+9=0\)
\(m=\frac{-(-4)+-\sqrt{(-4)^2-(4)(1)(9)}}{2}\)
\(=\frac{4\pm\sqrt{16-36}}{2}\)
\(=\frac{(4\pm i2\sqrt5)}{2}\)
\(m=2\pm i\sqrt5\)
Hence the general solution is
\(y=e^{2x}(A\cos\sqrt5x+B\sin\sqrt5x)\)
\(y(0)=0\)
\(\Rightarrow0=1(A(1)+B(0))\)
\(\Rightarrow A=0\)
\(y=e^{2x}(B\sin\sqrt5x)\)
\(y'(0)=-8\)
\(y'=2e^x(B\sin\sqrt5x)+\sqrt5e^x(B\cos\sqrt5x)\)
\(-8=0+\sqrt5(B)\)
\(B=\frac{-8}{\sqrt5}\)
Hence, the solution is
\(y=-e^{2x}(\frac{8}{\sqrt5}\sin\sqrt5x)\)

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