What is the solution for (dy)/(dx)=((3x^2-e^x))/(2y-5),y(0)=1 ?

The solution for (dy)/(dx)=((3x^2-e^x))/(2y-5),y(0)=1 is y=(5-sqrt(4x^3-4e^x+13))/2

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(dy)/(dx)=((3x^2-e^x))/(2y-5),y(0)=1

\frac{d y}{d x} = \frac{( 3 x ^{2} - e ^{x} )}{2 y - 5}, y (0) = 1

y = \frac{5 - 4 x ^{3} - 4 e ^{x} + 13}{2}

Solution steps

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

Solve separable ODE:

y = \frac{5 - 4 x ^{3} - 4 e ^{x} + 13}{2}

y = \frac{5 - 4 x ^{3} - 4 e ^{x} + 13}{2}

\frac{d y}{d x} = 10 - \frac{y}{50}

\frac{d y}{d x} = 5 x^{4} y^{2}

\frac{d y}{d x} x^{2} + x^{2} y^{2} + y^{2} = 0

\frac{d y}{d x} = \frac{( x y - 4 x + y - 4 )}{( x y - 6 x - 3 y + 18 )}

\frac{d y}{d t} = 16 y

What is the solution for (dy)/(dx)=((3x^2-e^x))/(2y-5),y(0)=1 ?
The solution for (dy)/(dx)=((3x^2-e^x))/(2y-5),y(0)=1 is y=(5-sqrt(4x^3-4e^x+13))/2