What is the solution for y^'=e^{4y+10x} ?

The solution for y^'=e^{4y+10x} is y=-(ln(-2/5 e^{10x}-c_{1}))/4

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y^'=e^{4y+10x}

y^{^{'}} = e^{4 y + 10 x}

y = - \frac{ln ( - \frac{2}{5} e ^{10 x} - c _{1} )}{4}

Solution steps

y^{'} (x) = e^{4 y + 10 x}

Solve separable ODE:

y = - \frac{ln ( - \frac{2}{5} e ^{10 x} - c _{1} )}{4}

y = - \frac{ln ( - \frac{2}{5} e ^{10 x} - c _{1} )}{4}

e^{(2 x)} y^{2} y^{^{'}} = 1

\frac{d y}{d x} + \frac{3 y}{x} = \frac{3}{x}

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

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

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

What is the solution for y^'=e^{4y+10x} ?
The solution for y^'=e^{4y+10x} is y=-(ln(-2/5 e^{10x}-c_{1}))/4