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

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

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(x^2+4)y^'+3xy=x

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

y = \frac{c _{1}}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{3}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{3}

y = \frac{c _{1}}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{3}

n = 1 \sum \infty \frac{7}{4 ^{n}}

x \to 0 lim (\frac{e ^{a x} - 1}{x})

x - a x

sin^{4} (a) cos^{9} (a)

\int 13 e^{- 13 x} d x

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