What is the solution for 4dy+ydx=x^2dy ?

The solution for 4dy+ydx=x^2dy is y=e^{-1/4 (ln(x/2+1)-ln(x/2-1))+c_{1}}

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4dy+ydx=x^2dy

4 d y + y d x = x^{2} d y

y = e^{- \frac{1}{4} (l n (\frac{x}{2} + 1) - l n (\frac{x}{2} - 1)) + c_{1}}

Solution steps

4 d y + y d x = x^{2} d y

Solve separable ODE:

y = e^{- \frac{1}{4} (l n (\frac{x}{2} + 1) - l n (\frac{x}{2} - 1)) + c_{1}}

y = e^{- \frac{1}{4} (l n (\frac{x}{2} + 1) - l n (\frac{x}{2} - 1)) + c_{1}}

y^{^{'}} = 2 x - 3 y + 1

e^{2 y} y^{^{'}} = 3

(90, 1), (980, 60)

n = 0 \sum \infty (\frac{x}{4})^{n}

\frac{d}{d x} (cos (x) - \frac{1}{6} cos (3 x))

What is the solution for 4dy+ydx=x^2dy ?
The solution for 4dy+ydx=x^2dy is y=e^{-1/4 (ln(x/2+1)-ln(x/2-1))+c_{1}}