What is the solution for y^'+y=-x^4 ?

The solution for y^'+y=-x^4 is y=-x^4+4x^3-12x^2+24x-24+(c_{1})/(e^x)

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y^'+y=-x^4

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

y = - x^{4} + 4 x^{3} - 12 x^{2} + 24 x - 24 + \frac{c _{1}}{e ^{x}}

Solution steps

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

Solve linear ODE:

y = - x^{4} + 4 x^{3} - 12 x^{2} + 24 x - 24 + \frac{c _{1}}{e ^{x}}

y = - x^{4} + 4 x^{3} - 12 x^{2} + 24 x - 24 + \frac{c _{1}}{e ^{x}}

\frac{\partial}{\partial y} (x + \frac{y}{z})

\int 7 x e^{6 x} d x

y = 2 x^{2} + 1, (3, 19)

f (x) = (1 + 2 x)^{2}, at x = 4

\frac{x ^{3}}{2 - x ^{2}}

What is the solution for y^'+y=-x^4 ?
The solution for y^'+y=-x^4 is y=-x^4+4x^3-12x^2+24x-24+(c_{1})/(e^x)