Question #294113

Solve the initial value problem





y"-3y'-10y=0, y'(0)=1, y'(0)=2

1
Expert's answer
2022-02-07T13:54:05-0500

y"3y10y=0,y(0)=1,y(0)=2y"-3y'-10y=0, y(0)=1, y'(0)=2

Auxiliary equation is

k23k10=0k^2-3k-10=0

k=3±9+402k=\frac{3\pm \sqrt{9+40}}{2}

k1=2,k2=5k_1=-2,k_2=5

General solution is

y=c1e2x+c2e5x...(1)y=c_1e^{-2x}+c_2e^{5x}...(1)

Since, y(0)=1y(0)=1

1=c1+c2...(2)1=c_1+c_2...(2)

And

y=2c1e2x+5c2e5xy'=-2c_1e^{-2x}+5c_2e^{5x}

Also y(0)=2y'(0)=2

2=2c1+5c2...(3)2=-2c_1+5c_2...(3)

Multiply equation (2) with 2 and add with equation (3), we get

4=7c2c2=474=7c_2\Rightarrow c_2=\frac{4}{7}

From equation (2),

1=c1+47c1=371=c_1+\frac 4 7\Rightarrow c_1=\frac 3 7


y=37e2x+47e5x7y=3e2x+4e5xy=\frac 3 7e^{-2x}+\frac 4 7e^{5x}\Rightarrow 7y=3e^{-2x}+4e^{5x}

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