Question

Solve the equation xdy/dx−ay=x+1, where a is a constant

Answer 1

Answer on Question #69632 – Math – Differential Equations

Question

Solve the equation

$$x \frac{dy}{dx} - ay = x + 1,$$

where $a$ is a constant.

Solution

A First Order Linear Nonhomogeneous DE can be solved with a help of the Method of Integrating Factor.

Divide both sides of (1) by $x$

$$\frac{dy}{dx} - \frac{a}{x}y = 1 + \frac{1}{x}$$

Integrating factor is

$$v(x) = e^{\int\left(-\frac{a}{x}\right)dx},$$

where

$$\int\left(-\frac{a}{x}\right)dx = -a\ln|x| + C.$$

Set the constant of integration $C$ to be equal to 0 in order to simplify $v$ and $|x| = x$ if $x > 0$.

Then

$$v(x) = e^{\int\left(-\frac{a}{x}\right)dx} = e^{-a\ln x} = \left(e^{\ln x}\right)^{-a} = x^{-a}$$

Multiple both sides of (2) by a positive function $v(x)$ that transforms the left-hand side into the derivative of the product $v(x) \cdot y$.

$$\begin{array}{l} x^{-a} \frac{dy}{dx} - x^{-a} \frac{a}{x}y = x^{-a}\left(1 + \frac{1}{x}\right) \\ \frac{d}{dx}(x^{-a}y) = x^{-a} + x^{-a-1} \end{array}$$

Case $a = 0$:

$$\begin{array}{l} \frac{dy}{dx} = 1 + \frac{1}{x} \\ dy = \left(1 + \frac{1}{x}\right)dx \\ \int dy = \int\left(1 + \frac{1}{x}\right)dx \\ y = x + \ln|x| + C_1 \end{array}$$

Case $a = 1$:

$$\frac{dy}{dx} - \frac{1}{x}y = 1 + \frac{1}{x}$$ $$\begin{array}{l} \frac{d}{dx}(x^{-1}y) = x^{-1} + x^{-1-1} \\ \int d(x^{-1}y) = \int \left(\frac{1}{x} + \frac{1}{x^2}\right) dx \\ \frac{y}{x} = \ln |x| - \frac{1}{x} + C_2 \\ y = x \ln |x| - 1 + C_2 x \\ \end{array}$$

Case $a \neq 0, a \neq 1$:

$$\begin{array}{l} \frac{d}{dx}(x^{-a}y) = x^{-a} + x^{-a-1} \\ \int (x^{-a}y) = \int (x^{-a} + x^{-a-1}) dx \\ x^{-a}y = \frac{x^{-a+1}}{-a+1} + \frac{x^{-a}}{-a} + C_3 \\ y = \frac{x}{-a+1} - \frac{1}{a} + C_3 x^a \\ \end{array}$$

Answer:

$$y = \frac{x}{-a+1} - \frac{1}{a} + C_3 x^a, \text{ if } a \neq 0, a \neq 1;$$ $$y = x + \ln |x| + C_1, \text{ if } a = 0;$$ $$y = x \ln |x| - 1 + C_2 x, \text{ if } a = 1.$$

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