What is the solution for e^xdx-ydy=0 ?

The solution for e^xdx-ydy=0 is y=sqrt(2e^x+c_{1)},y=-sqrt(2e^x+c_{1)}

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e^xdx-ydy=0

e^{x} d x - y d y = 0

y = 2 e^{x} + c_{1}, y = - 2 e^{x} + c_{1}

Solution steps

e^{x} d x - y d y = 0

Solve separable ODE:

y = 2 e^{x} + c_{1}, y = - 2 e^{x} + c_{1}

y = 2 e^{x} + c_{1}, y = - 2 e^{x} + c_{1}

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

y^{^{'}} + y sin (x) = sin (x)

(1 + x) y^{^{'}} - y = 0

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

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

What is the solution for e^xdx-ydy=0 ?
The solution for e^xdx-ydy=0 is y=sqrt(2e^x+c_{1)},y=-sqrt(2e^x+c_{1)}