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Popular >

y^'=e^{3y+4x}

Solution

y^{^{'}} = e^{3 y + 4 x}

Solution

y = - \frac{ln ( - \frac{3}{4} e ^{4 x} - c _{1} )}{3}

Solution steps

y^{'} (x) = e^{3 y + 4 x}

Solve separable ODE:

y = - \frac{ln ( - \frac{3}{4} e ^{4 x} - c _{1} )}{3}

y = - \frac{ln ( - \frac{3}{4} e ^{4 x} - c _{1} )}{3}

Popular Examples

e^x(dy)/(dx)=(x+xy^2)

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

y^'=(2x)/(1+2y),y(3)=0

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

(dy)/(dx)=(xy-x^3y)/(1-y)

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

xdy-ydx=0,y(1)=2

x d y - y d x = 0, y (1) = 2

(dy)/(dx)= y/x ,y(5)=20

\frac{d y}{d x} = \frac{y}{x}, y (5) = 20

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{3y+4x} ?
The solution for y^'=e^{3y+4x} is y=-(ln(-3/4 e^{4x}-c_{1}))/3