What is the solution for y^{''}+4y=2,e^xy(0)=2,y^'(0)=1 ?

The solution for y^{''}+4y=2,e^xy(0)=2,y^'(0)=1 is y=c_{1}cos(2x)+c_{2}sin(2x)+1/2

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y^{''}+4y=2,e^xy(0)=2,y^'(0)=1

y^{^{''}} + 4 y = 2, e^{x} y (0) = 2, y^{^{'}} (0) = 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{2}

Solution steps

y^{''} (x) + 4 y = 2

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{2}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{1}{2}

(d^{2} - d - 2) y = 10 cos (x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 4, e^{3} t y (0) = 2, y^{^{'}} (0) = 7

y^{^{''}} - 5 y^{^{'}} - 6 y = 6 x^{2} - 4 x - 3

(d^{2} + 4) y = sin^{2} (x)

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

What is the solution for y^{''}+4y=2,e^xy(0)=2,y^'(0)=1 ?
The solution for y^{''}+4y=2,e^xy(0)=2,y^'(0)=1 is y=c_{1}cos(2x)+c_{2}sin(2x)+1/2