# Differential Equations

1st Order Equations# Linear Differential Equations of First Order

Problems 1-2

Problems 3-7

### Definition of Linear Equation of First Order

A differental equation of type

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:

the integrating factor is defined by the formula

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:

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:

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 on problem description to see 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.*

Rewrite this equation in standard form:

We will solve this equation using the integrating factor

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:

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:

Then the derivative is given by

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 \({C\left( x \right)}:\)

where \({C_1}\) is an arbitrary real number.

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

Problems 1-2

Problems 3-7