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Popular >

dP-kPdt=0

Solution

d P - k P d t = 0

Solution

t = \frac{ln ( P )}{k} + c_{1}

Solution steps

d P - k P d t = 0

Let

t

be the dependent variable. Divide by

d P

:

1 - k P \frac{d t}{d P} = 0

Substitute

\frac{d t}{d P}

with

t^{'} (P)

1 - k P t^{^{'}} (P) = 0

Isolate

t^{'} (P) : t^{'} (P) = \frac{1}{k P}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{k P} d P

\int \frac{1}{k P} d P = \frac{1}{k} ln (P) + c_{1}

t = \frac{1}{k} ln (P) + c_{1}

Simplify

t = \frac{ln ( P )}{k} + c_{1}

Popular Examples

integral of 7sin(2t)

\int 7 sin (2 t) d t

area f(x)=x^2-4x,g(x)=0,x=0,x=4 area

f (x) = x^{2} - 4 x, g (x) = 0, x = 0, x = 4

(\partial)/(\partial y)(2y^{3/2})

\frac{\partial}{\partial y} (2 y^{\frac{3}{2}})

(y^'-e^{-t}+3)/y =-3,y(0)=4

\frac{y ^{^{'}} - e ^{- t} + 3}{y} = - 3, y (0) = 4

y^{''}-3y^'-4y=2sin(t)

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (t)

Frequently Asked Questions (FAQ)

What is the solution for dP-kPdt=0 ?
The solution for dP-kPdt=0 is t=(ln(P))/k+c_{1}