What is the solution for y^{''}-y^'=e^x(3sin(x)+cos(x)) ?

The solution for y^{''}-y^'=e^x(3sin(x)+cos(x)) is y=c_{1}e^x+c_{2}-e^xsin(x)-2e^xcos(x)

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y^{''}-y^'=e^x(3sin(x)+cos(x))

y^{^{''}} - y^{^{'}} = e^{x} (3 sin (x) + cos (x))

y = c_{1} e^{x} + c_{2} - e^{x} sin (x) - 2 e^{x} cos (x)

Solution steps

y^{''} (x) - y^{'} (x) = e^{x} (3 sin (x) + cos (x))

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - e^{x} sin (x) - 2 e^{x} cos (x)

y = c_{1} e^{x} + c_{2} - e^{x} sin (x) - 2 e^{x} cos (x)

4 \frac{d ^{2} y}{d x ^{2}} + 9 y = 15

y^{^{''}} - y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 4 x + 5

3 x^{^{''}} + 30 x^{^{'}} + 125 x = 350

y^{^{''}} + y = t^{2} + 1

What is the solution for y^{''}-y^'=e^x(3sin(x)+cos(x)) ?
The solution for y^{''}-y^'=e^x(3sin(x)+cos(x)) is y=c_{1}e^x+c_{2}-e^xsin(x)-2e^xcos(x)