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Popular >

y^'+2y-4=0

Solution

y^{^{'}} + 2 y - 4 = 0

Solution

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

Solution steps

y^{'} (t) + 2 y - 4 = 0

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)+ln(x)*y=3*ln(x)

\frac{d y}{d x} + ln (x) \cdot y = 3 \cdot ln (x)

(dx)/(dt)=14-(84/100)*x

\frac{d x}{d t} = 14 - (\frac{84}{100}) \cdot x

x + y x d x - d y = 0

(dy}{dx}=\frac{y^{1/2}ln(x))/x

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

(dy)/(dt)=k(y-2)

\frac{d y}{d t} = k (y - 2)

Frequently Asked Questions (FAQ)

What is the solution for y^'+2y-4=0 ?
The solution for y^'+2y-4=0 is y=e^{-2t-c_{1}}+2