⌨

You have entered
[src]

d --(x(t)) = x(t) dt

$$\frac{d}{d t} x{\left(t \right)} = x{\left(t \right)}$$

x' = x

Detail solution

Given the equation:

$$\frac{d}{d t} x{\left(t \right)} = x{\left(t \right)}$$

This differential equation has the form:

$$y' + P(x)y = 0$$,

where

$$P{\left(t \right)} = -1$$

and

and it is called

It's the equation with separable 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

$$P{\left(t \right)} = -1$$, then

$$\int P{\left(x \right)}\, dx = \int \left(-1\right)\, dt = - t + Const$$

Detailed solution of the integral

So, solution of the homogeneous linear equation:

$$y_{1} = e^{C_{1} + t}$$

$$y_{2} = - e^{C_{2} + t}$$

that leads to the correspondent solution for any constant C, not equal to zero:

$$y = C e^{t}$$

Graph of the Cauchy problem

The classification

1st exact

1st exact Integral

1st linear

1st linear Integral

1st power series

Bernoulli

Bernoulli Integral

almost linear

almost linear Integral

lie group

nth linear constant coeff homogeneous

separable

separable Integral

Numerical answer
[src]

(t, x):

(-10.0, 0.75)

(-7.777777777777778, 6.920861621442101)

(-5.555555555555555, 63.86442821962763)

(-3.333333333333333, nan)

(-1.1111111111111107, nan)

(1.1111111111111107, nan)

(3.333333333333334, nan)

(5.555555555555557, nan)

(7.777777777777779, nan)

(10.0, nan)

(10.0, nan)

The graph