What is the solution for y^{''}-2y^'+2y=e^x,y(pi/2)=0,y(0)=0 ?

The solution for y^{''}-2y^'+2y=e^x,y(pi/2)=0,y(0)=0 is y=e^x(-cos(x)-sin(x))+e^x

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y^{''}-2y^'+2y=e^x,y(pi/2)=0,y(0)=0

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x}, y (\frac{π}{2}) = 0, y (0) = 0

y = e^{x} (- cos (x) - sin (x)) + e^{x}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 2 y = e^{x}

Solve linear ODE:

y = e^{x} (- cos (x) - sin (x)) + e^{x}

y = e^{x} (- cos (x) - sin (x)) + e^{x}

What is the solution for y^{''}-2y^'+2y=e^x,y(pi/2)=0,y(0)=0 ?
The solution for y^{''}-2y^'+2y=e^x,y(pi/2)=0,y(0)=0 is y=e^x(-cos(x)-sin(x))+e^x