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dx/dt+x/t^2=1/t^2

Differential equation dx/dt+x/t^2=1/t^2

For Cauchy problem:

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

The graph:

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The solution

You have entered [src]
x(t)   d          1 
---- + --(x(t)) = --
  2    dt          2
 t                t 
$$\frac{d}{d t} x{\left(t \right)} + \frac{x{\left(t \right)}}{t^{2}} = \frac{1}{t^{2}}$$
x' + x/t^2 = t^(-2)
Detail solution
Given the equation:
$$\frac{d}{d t} x{\left(t \right)} + \frac{x{\left(t \right)}}{t^{2}} = \frac{1}{t^{2}}$$
This differential equation has the form:
y' + P(x)y = Q(x)

where
$$P{\left(t \right)} = \frac{1}{t^{2}}$$
and
$$Q{\left(t \right)} = \frac{1}{t^{2}}$$
and it is called linear homogeneous
differential first-order equation:
First of all, we should solve the correspondent linear homogeneous equation
y' + P(x)y = 0

with multiple variables
The equation is solved using following steps:
From y' + P(x)y = 0 you get

$$\frac{dy}{y} = - P{\left(x \right)} dx$$, if y is not equal to 0
$$\int \frac{1}{y}\, dy = - \int P{\left(x \right)}\, dx$$
$$\log{\left(\left|{y}\right| \right)} = - \int P{\left(x \right)}\, dx$$
Or,
$$\left|{y}\right| = e^{- \int P{\left(x \right)}\, dx}$$
Therefore,
$$y_{1} = e^{- \int P{\left(x \right)}\, dx}$$
$$y_{2} = - e^{- \int P{\left(x \right)}\, dx}$$
The expression indicates that it is necessary to find the integral:
$$\int P{\left(x \right)}\, dx$$
Because
$$P{\left(t \right)} = \frac{1}{t^{2}}$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int \frac{1}{t^{2}}\, dt = Const - \frac{1}{t}$$
Detailed solution of the integral
So, solution of the homogeneous linear equation:
$$y_{1} = e^{C_{1} + \frac{1}{t}}$$
$$y_{2} = - e^{C_{2} + \frac{1}{t}}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C e^{\frac{1}{t}}$$
We get a solution for the correspondent homogeneous equation
Now we should solve the inhomogeneous equation
y' + P(x)y = Q(x)

Use variation of parameters method
Now, consider C a function of x

$$y = C{\left(t \right)} e^{\frac{1}{t}}$$
And apply it in the original equation.
Using the rules
- for product differentiation;
- of composite functions derivative,
we find that
$$\frac{d}{d x} C{\left(x \right)} = Q{\left(x \right)} e^{\int P{\left(x \right)}\, dx}$$
Let use Q(x) and P(x) for this equation.
We get the first-order differential equation for C(x):
$$\frac{d}{d t} C{\left(t \right)} = \frac{e^{- \frac{1}{t}}}{t^{2}}$$
So, C(x) =
$$\int \frac{e^{- \frac{1}{t}}}{t^{2}}\, dt = Const + e^{- \frac{1}{t}}$$
Detailed solution of the integral
use C(x) at
$$y = C{\left(t \right)} e^{\frac{1}{t}}$$
and we get a definitive solution for y(x):
$$e^{\frac{1}{t}} \left(Const + e^{- \frac{1}{t}}\right)$$
The answer [src]
               1
               -
               t
x(t) = 1 + C1*e 
$$x{\left(t \right)} = C_{1} e^{\frac{1}{t}} + 1$$
The classification
separable
1st exact
1st linear
Bernoulli
almost linear
lie group
separable Integral
1st exact Integral
1st linear Integral
Bernoulli Integral
almost linear Integral
The graph
Differential equation dx/dt+x/t^2=1/t^2
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