how to solve differential equation y”+2y’-8y=0

Question: Solve differential equation y”+2y’-8y=0 , y(0)=1, y'(1)=1 Given homogeneous linear differential equation \[y”=-2y’+8y\] \[\ y(0)=1,\ y'(0)=1 \] \[y”+2y’-8y=0 \] (1) So it’s auxiliary equation is \[ \lambda^2+2\lambda-8=0\] \[\lambda^2+4\lambda-2\lambda-8=0\] \[\lambda\left(\lambda+4\right)-2\left(\lambda+4\right)=0\] \[\left(\lambda+4\right)\left(\lambda-2\right)=0\] It’s roots are \[ \lambda_1=-4 , \lambda_2=2\] Hence the…

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