What is the solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 ?

The solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 is y=cos(2x)+(sin(2x)(12-g))/8+(gx)/4

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y^{''}+4y=g(x),y(0)=1,y^'(0)=3

y^{^{''}} + 4 y = g (x), y (0) = 1, y^{^{'}} (0) = 3

y = cos (2 x) + \frac{sin ( 2 x ) ( 12 - g )}{8} + \frac{gx}{4}

Solution steps

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

Solve linear ODE:

y = cos (2 x) + \frac{sin ( 2 x ) ( 12 - g )}{8} + \frac{gx}{4}

y = cos (2 x) + \frac{sin ( 2 x ) ( 12 - g )}{8} + \frac{gx}{4}

What is the solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 ?
The solution for y^{''}+4y=g(x),y(0)=1,y^'(0)=3 is y=cos(2x)+(sin(2x)(12-g))/8+(gx)/4