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}

y^{^{'}} = y^{2} \cdot x

x y^{2} + y y^{^{'}} = x^{2} y y^{^{'}}

\frac{x ^{2}}{y ^{2} - 1} \frac{d y}{d x} = \frac{3 x ^{3}}{y}

\frac{d Q}{d t} + 0.05 Q = 0

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

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