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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + a y^{^{'}} + b y = x, y (0) = 0, y^{^{'}} (0) = 0

(D^{2} - 4 D + 3) y = 20 cos (x)

y^{^{''}} + (- 4) y^{^{'}} + 4 y = \frac{2 e ^{2}}{x}

y^{^{''}} - 9 y = 20 sin (x) + 3 x^{2}

y^{^{''}} + y^{^{'}} + y = k x

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