# Differential Equations

## First Order Equations # Linear Differential Equations of First Order

## Definition of Linear Equation of First Order

A differential equation of type

$y' + a\left( x \right)y = f\left( x \right),$

where $$a\left( x \right)$$ and $$f\left( x \right)$$ are continuous functions of $$x,$$ is called a linear nonhomogeneous differential equation of first order. We consider two methods of solving linear differential equations of first order:

• Using an integrating factor;
• Method of variation of a constant.

## Using an Integrating Factor

If a linear differential equation is written in the standard form:

$y' + a\left( x \right)y = f\left( x \right),$

the integrating factor is defined by the formula

$u\left( x \right) = \exp \left( {\int {a\left( x \right)dx} } \right).$

Multiplying the left side of the equation by the integrating factor $$u\left( x \right)$$ converts the left side into the derivative of the product $$y\left( x \right) u\left( x \right).$$

The general solution of the differential equation is expressed as follows:

$y = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}},$

where $$C$$ is an arbitrary constant.

## Method of Variation of a Constant

This method is similar to the previous approach. First it's necessary to find the general solution of the homogeneous equation:

$y' + a\left( x \right)y = 0.$

The general solution of the homogeneous equation contains a constant of integration $$C.$$ We replace the constant $$C$$ with a certain (still unknown) function $$C\left( x \right).$$ By substituting this solution into the nonhomogeneous differential equation, we can determine the function $$C\left( x \right).$$

The described algorithm is called the method of variation of a constant. Of course, both methods lead to the same solution.

## Initial Value Problem

If besides the differential equation, there is also an initial condition in the form of $$y\left( {{x_0}} \right) = {y_0},$$ such a problem is called the initial value problem (IVP) or Cauchy problem.

A particular solution for an IVP does not contain the constant $$C,$$ which is defined by substitution of the general solution into the initial condition $$y\left( {{x_0}} \right) = {y_0}.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the equation $y' - y - x{e^x} = 0.$

### Example 2

Solve the differential equation $xy' = y + 2{x^3}.$

### Example 1.

Solve the equation $y' - y - x{e^x} = 0.$

Solution.

We rewrite this equation in standard form:

$y' - y = x{e^x}.$

We will solve this equation using the integrating factor

$u\left( x \right) = {e^{\int {\left( { - 1} \right)dx} }} = {e^{ - \int {dx} }} = {e^{ - x}}.$

Then the general solution of the linear equation is given by

$y\left( x \right) = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} = \frac{{\int {\cancel{e^{ - x}}x\cancel{e^x}dx} + C}}{{{e^{ - x}}}} = \frac{{\int {xdx} + C}}{{{e^{ - x}}}} = {e^x}\left( {\frac{{{x^2}}}{2} + C} \right).$

### Example 2.

Solve the differential equation $xy' = y + 2{x^3}.$

Solution.

We will solve this problem by using the method of variation of a constant. First we find the general solution of the homogeneous equation:

$xy' = y,$

which can be solved by separating the variables:

$x\frac{{dy}}{{dx}} = y,\;\; \Rightarrow \frac{{dy}}{y} = \frac{{dx}}{x},\;\; \Rightarrow \int {\frac{{dy}}{y}} = \int {\frac{{dx}}{x}} ,\;\; \Rightarrow \ln \left| y \right| = \ln \left| x \right| + \ln C,\;\; \Rightarrow y = Cx,$

where $$C$$ is a positive real number.

Now we replace $$C$$ with a certain (still unknown) function $$C\left( x \right)$$ and will find a solution of the original nonhomogeneous equation in the form:

$y = C\left( x \right)x.$

Then the derivative is given by

$y' = {\left[ {C\left( x \right)x} \right]^\prime } = C'\left( x \right)x + C\left( x \right).$

Substituting this into the equation gives:

$x\left[ {C'\left( x \right)x + C\left( x \right)} \right] = C\left( x \right)x + 2{x^3},\;\; \Rightarrow C'\left( x \right){x^2} + \cancel{C\left( x \right)x} = \cancel{C\left( x \right)x} + 2{x^3},\;\; \Rightarrow C'\left( x \right) = 2x.$

Upon integration, we find the function $${C\left( x \right)}:$$

$C\left( x \right) = \int {2xdx} = {x^2} + {C_1},$

where $${C_1}$$ is an arbitrary real number.

Thus, the general solution of the given equation is written in the form

$y = C\left( x \right)x = \left( {{x^2} + {C_1}} \right)x = {x^3} + {C_1}x.$

See more problems on Page 2.