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Popular >

(dQ)/(dt)+0.05Q=0

Solution

\frac{d Q}{d t} + 0.05 Q = 0

Solution

Q = e^{- 0.05 t + c_{1}}

Solution steps

\frac{d Q}{d t} + 0.05 Q = 0

Solve separable ODE:

Q = e^{- 0.05 t + c_{1}}

Q = e^{- 0.05 t + c_{1}}

Popular Examples

y^'=(2+3x)sqrt(y),y(0)=3

y^{^{'}} = (2 + 3 x) y, y (0) = 3

y^'(s)=((1-2s)/(s^2))y(s)

y^{^{'}} (s) = (\frac{1 - 2 s}{s ^{2}}) y (s)

y^'+(cos(x))y=5cos(x)

y^{^{'}} + (cos (x)) y = 5 cos (x)

y^'= 3/2 y^{1/3},y(0)=0

y^{^{'}} = \frac{3}{2} y^{\frac{1}{3}}, y (0) = 0

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

x^{^{'}} = \frac{( x ^{2} + x )}{( 2 x + 1 )}

Frequently Asked Questions (FAQ)

What is the solution for (dQ)/(dt)+0.05Q=0 ?
The solution for (dQ)/(dt)+0.05Q=0 is Q=e^{-0.05t+c_{1}}