What is the solution for y^'+9xy=8x,y(0)=4 ?

The solution for y^'+9xy=8x,y(0)=4 is y=(28e^{-(9x^2)/2})/(9)+8/9

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y^'+9xy=8x,y(0)=4

y^{^{'}} + 9 x y = 8 x, y (0) = 4

y = \frac{28 e ^{- \frac{9 x ^{2}}{2}}}{9} + \frac{8}{9}

Solution steps

y^{'} (x) + 9 x y = 8 x

Solve separable ODE:

y = \frac{28 e ^{- \frac{9 x ^{2}}{2}}}{9} + \frac{8}{9}

y = \frac{28 e ^{- \frac{9 x ^{2}}{2}}}{9} + \frac{8}{9}

2 y^{^{'}} = e^{x - y}, y (0) = 0

x y^{^{'}} = y^{2} ln (x)

(x^{3} y^{3}) d x - 3 x y^{2} d y = 0

e^{x^{2}} \frac{d y}{d x} + 2 e^{x^{2}} x y = e^{x^{2}} x

d y = \frac{8 y}{x} d x

What is the solution for y^'+9xy=8x,y(0)=4 ?
The solution for y^'+9xy=8x,y(0)=4 is y=(28e^{-(9x^2)/2})/(9)+8/9