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Popular Calculus >

(dy)/(dt)=ky(t)^{1+c}

Solution

\frac{d y}{d t} = k y^{1 + c}

Solution

y = e^{\frac{k t ^{c + 2}}{c + 2} + c_{1}}

Solution steps

\frac{d y}{d t} = k y t^{1 + c}

Solve separable ODE:

y = e^{\frac{k t ^{c + 2}}{c + 2} + c_{1}}

y = e^{\frac{k t ^{c + 2}}{c + 2} + c_{1}}

Graph

Popular Examples

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y^'+1/(sqrt(1-x^2))y= 3/(sqrt(1-x^2))

(x+1)dx= 1/(-y^2)dy

y^'=((y^2+1))/((x^2+1))

Frequently Asked Questions (FAQ)

What is the solution for (dy)/(dt)=ky(t)^{1+c} ?
The solution for (dy)/(dt)=ky(t)^{1+c} is y=e^{(kt^{c+2})/(c+2)+c_{1}}