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 (x) and f (x) 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

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.