Mister Exam

# Differential equation dx/dt=ax-bx^2

y() =
y'() =
y''() =
y'''() =
y''''() =

from to

### The solution

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d                      2
--(x(t)) = a*x(t) - b*x (t)
dt                         
$$\frac{d}{d t} x{\left(t \right)} = a x{\left(t \right)} - b x^{2}{\left(t \right)}$$
x' = a*x - b*x^2
Detail solution
Given the equation:
$$\frac{d}{d t} x{\left(t \right)} = a x{\left(t \right)} - b x^{2}{\left(t \right)}$$
This differential equation has the form:
f1(x)*g1(x)*x' = f2(x)*g2(x),

where
$$\operatorname{f_{1}}{\left(t \right)} = 1$$
$$\operatorname{g_{1}}{\left(x \right)} = 1$$
$$\operatorname{f_{2}}{\left(t \right)} = 1$$
$$\operatorname{g_{2}}{\left(x \right)} = \left(a - b x{\left(t \right)}\right) x{\left(t \right)}$$
We give the equation to the form:
g1(x)/g2(x)*x'= f2(x)/f1(x).

Divide both parts of the equation by g2(x)
$$\left(a - b x{\left(t \right)}\right) x{\left(t \right)}$$
we get
$$\frac{\frac{d}{d t} x{\left(t \right)}}{\left(a - b x{\left(t \right)}\right) x{\left(t \right)}} = 1$$
We separated the variables t and x.

Now, multiply the both equation sides by dt,
then the equation will be the
$$\frac{dt \frac{d}{d t} x{\left(t \right)}}{\left(a - b x{\left(t \right)}\right) x{\left(t \right)}} = dt$$
or
$$\frac{dx}{\left(a - b x{\left(t \right)}\right) x{\left(t \right)}} = dt$$

Take the integrals from the both equation sides:
- the integral of the left side by x,
- the integral of the right side by t.
$$\int \frac{1}{x \left(a - b x\right)}\, dx = \int 1\, dt$$
Detailed solution of the integral with x
Detailed solution of the integral with t
Take this integrals
$$- \frac{- \log{\left(x \right)} + \log{\left(- \frac{a}{b} + x \right)}}{a} = Const + t$$
Detailed solution of the equation
We get the simple equation with the unknown variable x.
(Const - it is a constant)

Solution is:
$$\operatorname{x_{1}} = x{\left(t \right)} = \frac{a e^{a \left(C_{1} + t\right)}}{b \left(e^{a \left(C_{1} + t\right)} - 1\right)}$$
             a*(C1 + t)
a*e
x(t) = --------------------
/      a*(C1 + t)\
b*\-1 + e          /
$$x{\left(t \right)} = \frac{a e^{a \left(C_{1} + t\right)}}{b \left(e^{a \left(C_{1} + t\right)} - 1\right)}$$
The classification
separable
1st power series
lie group
separable Integral
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