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Popular >

y^'=-5y+20

Solution

y^{^{'}} = - 5 y + 20

Solution

y = e^{- 5 t - c_{1}} + 4

Solution steps

y^{'} (t) = - 5 y + 20

Solve separable ODE:

y = e^{- 5 t - c_{1}} + 4

y = e^{- 5 t - c_{1}} + 4

Graph

Popular Examples

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

\frac{d y}{d x} y = \frac{1}{1 - x ^{2}}

2xy+(1+x^2)*y^'=0

2 x y + (1 + x^{2}) \cdot y^{^{'}} = 0

y^'=8tcos^2(y),y(1)=(3pi)/4

y^{^{'}} = 8 t cos^{2} (y), y (1) = \frac{3 π}{4}

derivative of (x^8/(y^5))=0

\frac{d}{d x} (\frac{x ^{8}}{y ^{5}}) = 0

y^{^{'}} = (\frac{1}{1 + x}) y

Frequently Asked Questions (FAQ)

What is the solution for y^'=-5y+20 ?
The solution for y^'=-5y+20 is y=e^{-5t-c_{1}}+4