What is the solution for xy^'+4y=-25xsin(x^5) ?

The solution for xy^'+4y=-25xsin(x^5) is y=(5cos(x^5))/(x^4)+(c_{1})/(x^4)

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xy^'+4y=-25xsin(x^5)

x y^{^{'}} + 4 y = - 25 x sin (x^{5})

y = \frac{5 cos ( x ^{5} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

x y^{'} (x) + 4 y = - 25 x sin (x^{5})

Solve linear ODE:

y = \frac{5 cos ( x ^{5} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

y = \frac{5 cos ( x ^{5} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

What is the solution for xy^'+4y=-25xsin(x^5) ?
The solution for xy^'+4y=-25xsin(x^5) is y=(5cos(x^5))/(x^4)+(c_{1})/(x^4)