What is the solution for (y^')/x =e^{y+x^2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y^')/x =e^{y+x^2}

\frac{y ^{^{'}}}{x} = e^{y + x^{2}}

y = - ln (- \frac{1}{2} e^{x^{2}} - c_{1})

Solution steps

\frac{y ^{'} ( x )}{x} = e^{y + x^{2}}

Solve separable ODE:

y = - ln (- \frac{1}{2} e^{x^{2}} - c_{1})

y = - ln (- \frac{1}{2} e^{x^{2}} - c_{1})

\frac{d y}{d x} + \frac{x y}{x ^{2} - 9} = 0

16 y y^{^{'}} + 9 x = 0

x y^{^{'}} = y + 1 + x y + x

y^{^{'}} = \frac{sin ( 2 x )}{cos ( 3 y )}, y (0) = 1

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 4

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