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Popular >

y^'+5e^{x+y}=0

Solution

y^{^{'}} + 5 e^{x + y} = 0

Solution

y = - ln (5 e^{x} - c_{1})

Solution steps

y^{'} (x) + 5 e^{x + y} = 0

Solve separable ODE:

y = - ln (5 e^{x} - c_{1})

y = - ln (5 e^{x} - c_{1})

Popular Examples

derivative of f(x)=(sqrt(x))/(4+x)derivative of

f (x) = \frac{x}{4 + x}

integral of 1/((x+1)(x^2+1))

\int \frac{1}{( x + 1 ) ( x ^{2} + 1 )} d x

integral of x^3sqrt(x^2-9)

\int x^{3} x^{2} - 9 d x

(dy)/(dx)=2^{x-y},y(-3)=-5

\frac{d y}{d x} = 2^{x - y}, y (- 3) = - 5

integral of (4x+8)/((x+5)^2)

\int \frac{4 x + 8}{( x + 5 ) ^{2}} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'+5e^{x+y}=0 ?
The solution for y^'+5e^{x+y}=0 is y=-ln(5e^x-c_{1})