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Popular >

y^'+k^2y=0

Solution

y^{^{'}} + k^{2} y = 0

Solution

y = e^{- k^{2} t + c_{1}}

Solution steps

y^{'} (t) + k^{2} y = 0

Solve separable ODE:

y = e^{- k^{2} t + c_{1}}

y = e^{- k^{2} t + c_{1}}

Popular Examples

(1+x^3)(dy)/(dx)=x^2y

(1 + x^{3}) \frac{d y}{d x} = x^{2} y

(dy)/(dx)=(x(3+y^2))/(y(2+x^2))

\frac{d y}{d x} = \frac{x ( 3 + y ^{2} )}{y ( 2 + x ^{2} )}

(dy)/(dt)=2sqrt(y)sqrt(t)

\frac{d y}{d t} = 2 y t

8 N^{^{'}} + 10 t N^{5} = 0

4sqrt(xy)(dy)/(dx)=1,y>0

4 x y \frac{d y}{d x} = 1, y > 0

Frequently Asked Questions (FAQ)

What is the solution for y^'+k^2y=0 ?
The solution for y^'+k^2y=0 is y=e^{-k^2t+c_{1}}