dy/dx-3y=6

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d -3*y(x) + --(y(x)) = 6 dx

$$- 3 y{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} = 6$$

-3*y + y' = 6

Detail solution

Given the equation:

$$- 3 y{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} = 6$$

This differential equation has the form:

where

$$P{\left(x \right)} = -3$$

and

$$Q{\left(x \right)} = 6$$

and it is called**linear inhomogeneous**

**differential first-order equation:**

First of all, we should solve the correspondent linear homogeneous equation

with multiple variables

The equation is solved using following steps:

$$\frac{dy}{y} = - P{\left(x \right)} dx$$, 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

$$P{\left(x \right)} = -3$$, then

$$\int P{\left(x \right)}\, dx = \int \left(-3\right)\, dx = - 3 x + Const$$

Detailed solution of the integral

So, solution of the homogeneous linear equation:

$$y_{1} = e^{C_{1} + 3 x}$$

$$y_{2} = - e^{C_{2} + 3 x}$$

that leads to the correspondent solution

for any constant C, not equal to zero:

$$y = C e^{3 x}$$

We get a solution for the correspondent homogeneous equation

Now we should solve the inhomogeneous equation

Use variation of parameters method

Now, consider C a function of x

$$y = C{\left(x \right)} e^{3 x}$$

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):

$$\frac{d}{d x} C{\left(x \right)} = 6 e^{- 3 x}$$

So,

$$C{\left(x \right)} = \int 6 e^{- 3 x}\, dx = Const - 2 e^{- 3 x}$$

Detailed solution of the integral

use C(x) at

$$y = C{\left(x \right)} e^{3 x}$$

and we get a definitive solution for y(x):

$$e^{3 x} \left(Const - 2 e^{- 3 x}\right)$$

$$- 3 y{\left(x \right)} + \frac{d}{d x} y{\left(x \right)} = 6$$

This differential equation has the form:

y' + P(x)y = Q(x)

where

$$P{\left(x \right)} = -3$$

and

$$Q{\left(x \right)} = 6$$

and it is called

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

$$P{\left(x \right)} = -3$$, then

$$\int P{\left(x \right)}\, dx = \int \left(-3\right)\, dx = - 3 x + Const$$

Detailed solution of the integral

So, solution of the homogeneous linear equation:

$$y_{1} = e^{C_{1} + 3 x}$$

$$y_{2} = - e^{C_{2} + 3 x}$$

that leads to the correspondent solution

for any constant C, not equal to zero:

$$y = C e^{3 x}$$

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

$$y = C{\left(x \right)} e^{3 x}$$

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):

$$\frac{d}{d x} C{\left(x \right)} = 6 e^{- 3 x}$$

So,

$$C{\left(x \right)} = \int 6 e^{- 3 x}\, dx = Const - 2 e^{- 3 x}$$

Detailed solution of the integral

use C(x) at

$$y = C{\left(x \right)} e^{3 x}$$

and we get a definitive solution for y(x):

$$e^{3 x} \left(Const - 2 e^{- 3 x}\right)$$

The answer (#2)
[src]

$$y\left(x\right)=\left(y\left(0\right)+2\right)\,e^{3\,x}-2$$

y = (y(0)+2)*E^(3*x)-2

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 undetermined coefficients

nth linear constant coeff variation of parameters

nth linear constant coeff variation of parameters Integral

separable

separable Integral

Numerical answer
[src]

(x, y):

(-10.0, 0.75)

(-7.777777777777778, 2158.8732009891255)

(-5.555555555555555, 1697951.835945232)

(-3.333333333333333, 1334204720.233111)

(-1.1111111111111107, 1048380823475.5973)

(1.1111111111111107, 823788383194792.0)

(3.333333333333334, 6.47309913607928e+17)

(5.555555555555557, 5.08638059007202e+20)

(7.777777777777779, 3.9967358700319505e+23)

(10.0, 3.1405234689327076e+26)

(10.0, 3.1405234689327076e+26)