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

The solution for y^{''}+4y=2x,y(0)=1,y^'(0)=2 is y=cos(2x)+3/4 sin(2x)+x/2

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

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

y = cos (2 x) + \frac{3}{4} sin (2 x) + \frac{x}{2}

Solution steps

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

Solve linear ODE:

y = cos (2 x) + \frac{3}{4} sin (2 x) + \frac{x}{2}

y = cos (2 x) + \frac{3}{4} sin (2 x) + \frac{x}{2}

\frac{d ^{2} u}{d x ^{2}} - u = 6

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = e^{x}

y^{^{''}} - 4 y = - 7 e^{2 x} + x

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + t

y^{^{''}} + y^{^{'}} = cos (2 t)

What is the solution for y^{''}+4y=2x,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+4y=2x,y(0)=1,y^'(0)=2 is y=cos(2x)+3/4 sin(2x)+x/2