What is the solution for 4x^3y+x^4y^'=sin^3(x/9) ?

The solution for 4x^3y+x^4y^'=sin^3(x/9) is y=(3(cos^3(x/9)-3cos(x/9)))/(x^4)+(c_{1})/(x^4)

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4x^3y+x^4y^'=sin^3(x/9)

4 x^{3} y + x^{4} y^{^{'}} = sin^{3} (\frac{x}{9})

y = \frac{3 ( cos ^{3} ( \frac{x}{9} ) - 3 cos ( \frac{x}{9} ) )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

4 x^{3} y + x^{4} y^{'} (x) = sin^{3} (\frac{x}{9})

Solve linear ODE:

y = \frac{3 ( cos ^{3} ( \frac{x}{9} ) - 3 cos ( \frac{x}{9} ) )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

y = \frac{3 ( cos ^{3} ( \frac{x}{9} ) - 3 cos ( \frac{x}{9} ) )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

ln (x + 1), 0

\frac{d}{d x} (3 sin (2 x) cos (x))

\frac{\partial}{\partial t} (1)

f (x) = \frac{x ^{2} + 2 x + 3}{x ^{2} + 2 x + 1}

\frac{\partial}{\partial x} (- yz)

What is the solution for 4x^3y+x^4y^'=sin^3(x/9) ?
The solution for 4x^3y+x^4y^'=sin^3(x/9) is y=(3(cos^3(x/9)-3cos(x/9)))/(x^4)+(c_{1})/(x^4)