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}}

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)