# 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 3

Solve the equation $$y’ – 2y = x.$$

### Example 4

Solve the differential equation $${x^2}y’ + xy + 2$$ $$= 0.$$

### Example 5

Solve the initial value problem: $$y’ – y\tan x$$ $$= \sin x,$$ $$y\left( 0 \right) = 1.$$

### Example 6

Solve the differential equation (IVP) $$y’ + {\large\frac{3}{x}\normalsize}y$$ $$= {\large\frac{2}{{{x^2}}}\normalsize}$$ with the initial condition $$y\left( 1 \right) = 2.$$

### Example 7

Find the general solution of the differential equation $$y = \left( {2{y^4} + 2x} \right)y’.$$

### 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

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

### 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

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

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Problems 1-2
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Problems 3-7