Differential Equations

First Order Equations

1st Order Differential Equations Logo

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 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

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

    Page 1
    Problems 1-2
    Page 2
    Problems 3-7