What is the solution for xy^'+4y=9x^5+2x^3 ?

The solution for xy^'+4y=9x^5+2x^3 is y=x^5+(2x^3)/7+(c_{1})/(x^4)

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xy^'+4y=9x^5+2x^3

x y^{^{'}} + 4 y = 9 x^{5} + 2 x^{3}

y = x^{5} + \frac{2 x ^{3}}{7} + \frac{c _{1}}{x ^{4}}

Solution steps

x y^{'} (x) + 4 y = 9 x^{5} + 2 x^{3}

Solve linear ODE:

y = x^{5} + \frac{2 x ^{3}}{7} + \frac{c _{1}}{x ^{4}}

y = x^{5} + \frac{2 x ^{3}}{7} + \frac{c _{1}}{x ^{4}}

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 t}

\int \frac{1}{5 + 2 t} d t

\frac{\partial}{\partial a} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

2 (x - y) d x + d y = 0

x \to 1 + lim ((x)^{\frac{1}{1 - x}})

What is the solution for xy^'+4y=9x^5+2x^3 ?
The solution for xy^'+4y=9x^5+2x^3 is y=x^5+(2x^3)/7+(c_{1})/(x^4)