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

The solution for y^{''}+4=xcos(x) is y=-xcos(x)+2sin(x)-2x^2+c_{1}x+c_{2}

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

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

y = - x cos (x) + 2 sin (x) - 2 x^{2} + c_{1} x + c_{2}

Solution steps

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

Isolate

y^{''} (x) : y^{''} (x) = x cos (x) - 4

Solve

y^{''} (x) = x cos (x) - 4 : y^{'} (x) = x sin (x) + cos (x) - 4 x + c_{1}

Solve

y^{'} (x) = x sin (x) + cos (x) - 4 x + c_{1} : y = - x cos (x) + 2 sin (x) - 2 x^{2} + c_{1} x + c_{2}

y = - x cos (x) + 2 sin (x) - 2 x^{2} + c_{1} x + c_{2}

y^{^{'}} = - 5 \frac{x}{2} - \frac{2}{5} x

\frac{x ( 1 - x ) y ^{^{''}} - 5}{2 y ^{^{'}} - y} = 0

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

f^{^{'}} (x) = 3 x^{2} + 8 x + 2

f^{^{'}} (x) = \frac{2 cos ^{2} ( x )}{a ^{9}}

What is the solution for y^{''}+4=xcos(x) ?
The solution for y^{''}+4=xcos(x) is y=-xcos(x)+2sin(x)-2x^2+c_{1}x+c_{2}