Question

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

a. y=x/1−a − 1/a+cx
b. y=x/1−a + 1/a+cx
c. y=x/1+a − 1/a+cx
d. y=−x/1−a − 1/a+cx

Answer 1

Answer on Question #79776 – Math – Differential Equations

Question

Solve the equation

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

where $a$ is a constant.

$$a) \quad y = \frac{x}{1 - a} - \frac{1}{a} + Cx;$$ $$b) \quad y = \frac{x}{1 - a} + \frac{1}{a} + Cx;$$ $$c) \quad y = \frac{x}{1 + a} - \frac{1}{a} + Cx;$$ $$d) \quad y = - \frac{x}{1 - a} - \frac{1}{a} + Cx.$$

Solution

Let's write the equation in the form:

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

It's a first order linear differential equation, in turn that means it looks like

$$\frac{dy}{dx} + P(x) y = Q(x)$$

with $P(x) = -\frac{a}{x}$ and $Q(x) = 1 + \frac{1}{x}$.

It can be integrated in three ways.

1) The first way is using auxiliary function.

Let

$$y = uv,$$

where $u = u(x)$ and $v = v(x)$ are the auxiliary functions.

Then

$$\frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx}.$$

So we have:

$$\frac{du}{dx} v + u \frac{dv}{dx} - \frac{a}{x} uv = 1 + \frac{1}{x}$$

or

$$\frac {d u}{d x} v + u \left(\frac {d v}{d x} - \frac {a}{x} v\right) = 1 + \frac {1}{x}.$$

Since one of the auxiliary functions can be chosen arbitrarily, let's choose a function $v$ such that it will satisfy the condition:

$$\frac {d v}{d x} - \frac {a}{x} v = 0.$$

So we have

$$\frac {d v}{d x} = \frac {a}{x} v.$$

It's a separable equation what can be easily integrated:

$$\int \frac {d v}{v} = a \int \frac {d x}{x};$$ $$\ln | v | = a \ln | x |$$ $$v = x ^ {a}.$$

Substitute the expression for the function $v$ in equation:

$$\frac {d u}{d x} x ^ {a} = 1 + \frac {1}{x}$$

or

$$\frac {d u}{d x} = \frac {1}{x ^ {a}} + \frac {1}{x ^ {a + 1}}.$$

It's a separable equation so we can integrate it:

$$\int d u = \int \left(\frac {1}{x ^ {a}} + \frac {1}{x ^ {a + 1}}\right) d x;$$ $$u = \frac {x ^ {1 - a}}{1 - a} - \frac {x ^ {- a}}{a} + C = \frac {x}{(1 - a) x ^ {a}} - \frac {1}{a x ^ {a}} + C, \quad \text{where } C = \text{const}.$$

After reverse replacement we have a general solution of the given equation:

$$y = \frac {x}{1 - a} - \frac {1}{a} + C x ^ {a}, \quad \text{where } C = \text{const}.$$

2) The second way is a use of an integrating factor

$$\mu (x) = e ^ {\int P (x) d x}.$$

The solution is then commonly written as

$$y = \frac {1}{\mu (x)} \int \mu (x) Q (x) d x.$$

Determine the integrating factor:

$$\mu (x) = e ^ {- a \int \frac {d x}{x}} = e ^ {- a \ln | x |} = x ^ {- a}.$$

Then the solution is

$$y = x^a \int x^{-a} \left(1 + \frac{1}{x}\right) dx = x^a \int \left(\frac{1}{x^a} + \frac{1}{x^{a+1}}\right) dx = x^a \left(\frac{x^{1-a}}{1-a} - \frac{x^{-a}}{a} + C\right) = \frac{x}{1-a} - \frac{1}{a} + C x^a.$$

So the general solution of the given equation is

$$y = \frac{x}{1-a} - \frac{1}{a} + C x^a, \quad \text{where } C = \text{const}.$$

3) The third way is a use of the variation of constants method.

We start with the homogeneous equation:

$$\frac{dy}{dx} - \frac{a}{x} y = 0,$$

which is a separable equation, so we can integrate it:

$$\int \frac{dy}{y} = a \int \frac{dx}{x};$$ $$\ln |y| = a \ln |x| + \ln C;$$ $$y = C x^a.$$

The idea of the variation of constants method is to replace constant $C$ by a function $C(x)$. So we have:

$$y = C(x) x^a$$

and

$$\frac{dy}{dx} = \frac{dC}{dx} x^a + C a x^{a-1}.$$

Substitute the expression for $y$ and $\frac{dy}{dx}$ in the equation:

$$x \left(\frac{dC}{dx} x^a + C a x^{a-1}\right) - a C x^a = x + 1;$$ $$\frac{dC}{dx} x^{a+1} + C a x^a - a C x^a = x + 1;$$ $$\frac{dC}{dx} x^{a+1} = x + 1.$$

The obtained equation is a separable equation and can be easily integrated:

$$C(x) = \int \left(\frac{1}{x^a} + \frac{1}{x^{a+1}}\right) dx = \frac{x^{1-a}}{1-a} - \frac{x^{-a}}{a} + C_1,$$

and the general solution of the given equation is:

$$y = \left(\frac{x^{1-a}}{1-a} - \frac{x^{-a}}{a} + C_1\right) x^a = \frac{x}{1-a} - \frac{1}{a} + C_1 x^a.$$

The result is evidently the same as what we found using a substitution (an auxiliary function) or an integrating factor.

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