What is the solution for y^{''}+16y=7x,y(0)=1,y^'(0)=4 ?

The solution for y^{''}+16y=7x,y(0)=1,y^'(0)=4 is y=cos(4x)+57/64 sin(4x)+(7x)/(16)

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y^{''}+16y=7x,y(0)=1,y^'(0)=4

y^{^{''}} + 16 y = 7 x, y (0) = 1, y^{^{'}} (0) = 4

y = cos (4 x) + \frac{57}{64} sin (4 x) + \frac{7 x}{16}

Solution steps

y^{''} (x) + 16 y = 7 x

Solve linear ODE:

y = cos (4 x) + \frac{57}{64} sin (4 x) + \frac{7 x}{16}

y = cos (4 x) + \frac{57}{64} sin (4 x) + \frac{7 x}{16}

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 cos (x)

y^{^{''}} + 4 y = 3 x (2 + cos (2 x))

y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 2 = 0

What is the solution for y^{''}+16y=7x,y(0)=1,y^'(0)=4 ?
The solution for y^{''}+16y=7x,y(0)=1,y^'(0)=4 is y=cos(4x)+57/64 sin(4x)+(7x)/(16)