# How to solve differential linear equations

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A differential equation in which an unknown function and its derivative enter linearly, that is, to the first power, is called a linear first-order differential equation.

Instructions

- 1

The general form of the linear differential equation of the first order is as follows:

Y? + P (x) * y = f (x),

Where y is an unknown function, and p (x) and f (x) -Some given functions. They are considered continuous in the region in which it is required to integrate the equation. In particular, they can be constants.

- 2

If f (x)? 0, then the equation is said to be homogeneous, if not, then, respectively, is inhomogeneous.

- 3

A linear homogeneous equation can be solved by the method of separation of variables. Its general form: y? + P (x) * y = 0, therefore:

Dy / dx = -p (x) * y, which implies that dy / y = -p (x) dx.

- 4

Integrating both sides of the resulting equality, we obtain:

(Dy / y) = -p (x) dx, that is, ln (y) = -p (x) dx + ln (C) or y = C * e ^ (-? P (x) dx) ).

- 5

The solution of the inhomogeneous linear equation can beDerive from the solution of the corresponding homogeneous, that is, the same equation with the discarded right-hand side f (x). To do this, we must replace the constant C in the solution of the homogeneous equation by the unknown function f (x). Then the solution of the inhomogeneous equation will be represented in the form:

Y =? (X) * e ^ (-? P (x) dx)).

- 6

Differentiating this expression, we get that the derivative of y is equal to:

Y? =? (X) * e ^ (-? P (x) dx) -? (X) * p (x) * e ^ (-? P (x) dx).

Substituting the expressions for y and y? Into the initial equation and simplifying the result, it is easy to arrive at the result:

D? / Dx = f (x) * e ^ (? P (x) dx).

- 7

After integrating both sides of the equation, it becomes:

? (X) =? (F (x) * e ^ (? P (x) dx)) dx + C1.

Thus, the desired function y is expressed as:

Y = e ^ (-? P (x) dx) * (C +? F (x) * e ^ (? P (x) dx)) dx).

- 8

If we equate the constant C with zero, then from the expression for y we can obtain a particular solution of the given equation:

Y1 = (e ^ (-? P (x) dx)) * (? F (x) * e ^ (? P (x) dx)) dx).

Then the complete solution can be expressed in the form:

Y = y1 + C * e ^ (-? P (x) dx)).

- 9

In other words, the complete solution of the linearInhomogeneous differential equation of the first order is equal to the sum of its particular solution and the general solution of the corresponding homogeneous first-order linear equation.