What is the solution for (x^2+4)y^'+3xy=x,y(0)=1 ?

The solution for (x^2+4)y^'+3xy=x,y(0)=1 is y=(16)/(3(x^2+4)^{3/2)}+1/3

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(x^2+4)y^'+3xy=x,y(0)=1

(x^{2} + 4) y^{^{'}} + 3 x y = x, y (0) = 1

y = \frac{16}{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{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{3}

y = \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}} + \frac{1}{3}

9 x^{4} - 9 x^{2}, 8 x^{2}

\int_{0}^{\frac{1}{10}} \frac{1}{( 100 x ^{2} + 1 ) ^{\frac{3}{2}}} d x

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\frac{d y}{d x}, 9 x y = 5

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What is the solution for (x^2+4)y^'+3xy=x,y(0)=1 ?
The solution for (x^2+4)y^'+3xy=x,y(0)=1 is y=(16)/(3(x^2+4)^{3/2)}+1/3