What is the solution for y(2x^2+5x-3)dy=e^ydx ?

The solution for y(2x^2+5x-3)dy=e^ydx is y=-\W((c_{1}+ln(2x-1)-ln(x+3))/(7e))-1

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y(2x^2+5x-3)dy=e^ydx

y (2 x^{2} + 5 x - 3) d y = e^{y} d x

y = - W (\frac{c _{1} + ln ( 2 x - 1 ) - ln ( x + 3 )}{7 e}) - 1

Solution steps

y (2 x^{2} + 5 x - 3) d y = e^{y} d x

Solve separable ODE:

y = - W (\frac{c _{1} + ln ( 2 x - 1 ) - ln ( x + 3 )}{7 e}) - 1

y = - W (\frac{c _{1} + ln ( 2 x - 1 ) - ln ( x + 3 )}{7 e}) - 1

\frac{\partial}{\partial x} (x e^{7 x y})

\int (x^{2} + 1) 2 x + 1 d x

\frac{( 3 s e ^{(- 2 s)} )}{4 s ^{2} + 36}

\int x^{2} cos (x) - 2 x cos (x) d x

x \to 9 lim (\frac{x - 9}{∣ x - 9 ∣})

What is the solution for y(2x^2+5x-3)dy=e^ydx ?
The solution for y(2x^2+5x-3)dy=e^ydx is y=-\W((c_{1}+ln(2x-1)-ln(x+3))/(7e))-1