{"id":835,"date":"2025-06-20T17:16:35","date_gmt":"2025-06-20T17:16:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=835"},"modified":"2025-09-10T17:29:06","modified_gmt":"2025-09-10T17:29:06","slug":"first-order-linear-equations-and-applications-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-learn-it-3\/","title":{"raw":"First-Order Linear Equations and Applications: Learn It 3","rendered":"First-Order Linear Equations and Applications: Learn It 3"},"content":{"raw":"

Integrating Factors<\/h2>\r\n

We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:<\/p>\r\n\r\n

[latex]y^{\\prime} +p\\left(x\\right)y=q\\left(x\\right)[\/latex].<\/div>\r\n

The first term on the left-hand side is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the product rule from differentiation.<\/p>\r\n

If we multiply [latex]y' +p(x)y=q(x)[\/latex] by a yet-to-be-determined function [latex]\\mu(x)[\/latex], then the equation becomes:<\/p>\r\n\r\n

[latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\r\n

The left-hand side of [latex]y^{\\prime} +p\\left(x\\right)y=q\\left(x\\right)[\/latex] can be matched perfectly to the product rule.<\/p>\r\n\r\n

\r\n

Recall the product rule:<\/p>\r\n

[latex]\\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)[\/latex]<\/p>\r\n\r\n<\/section>\r\n

Matching term by term gives us [latex]y=f(x)[\/latex], [latex]g(x)=\\mu(x)[\/latex], and [latex]g'(x)=\\mu(x)p(x)[\/latex]. Taking the derivative of [latex]g(x)=\\mu(x)[\/latex] and setting it equal to the right-hand side of [latex]g'(x)=\\mu(x)p(x)[\/latex] leads to:<\/p>\r\n\r\n

[latex]{\\mu }^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex].<\/div>\r\n

This is a first-order, separable differential equation for [latex]\\mu(x)[\/latex]. We know [latex]p(x)[\/latex] because it appears in the differential equation we are solving. Separating variables and integrating yields:<\/p>\r\n\r\n

[latex]\\begin{array}{ccc}\\hfill \\frac{{\\mu }^{\\prime }\\left(x\\right)}{\\mu \\left(x\\right)}& =\\hfill & p\\left(x\\right)\\hfill \\\\ \\hfill {\\displaystyle\\int \\frac{{\\mu }^{\\prime }\\left(x\\right)}{\\mu \\left(x\\right)}dx}& =\\hfill & {\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\\\ \\hfill \\text{ln}|\\mu \\left(x\\right)|& =\\hfill & {\\displaystyle\\int p\\left(x\\right)dx+C}\\hfill \\\\ \\hfill {e}^{\\text{ln}|\\mu \\left(x\\right)|}& =\\hfill & {e}^{\\displaystyle\\int p\\left(x\\right)dx+C}\\hfill \\\\ \\hfill |\\mu \\left(x\\right)|& =\\hfill & {C}_{1}{e}^{\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\\\ \\hfill \\mu \\left(x\\right)& =\\hfill & {C}_{2}{e}^{\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\end{array}[\/latex]<\/div>\r\n

Here [latex]C_2[\/latex] can be an arbitrary (positive or negative) constant. Since we only need one integrating factor to solve our equation, we can set [latex]C_2 = 1[\/latex]. This leads to a general method for solving a first-order linear differential equation.<\/p>\r\n

We first multiply both sides of [latex]y' +p(x)y=q(x)[\/latex] by the integrating factor [latex]\\mu(x)[\/latex]. This gives:<\/p>\r\n

[latex]\\mu(x)y' +\\mu(x)p(x)y=\\mu(x)q(x)[\/latex]<\/p>\r\n

The left-hand side can be rewritten as [latex]\\frac{d}{dx}(\\mu(x)y)[\/latex]:<\/p>\r\n

[latex]\\frac{d}{dx}(\\mu(x)y)=\\mu(x)q(x)[\/latex]<\/p>\r\n

Next integrate both sides with respect to [latex]x[\/latex]:<\/p>\r\n

[latex]\\begin{array}{ccc}\\hfill \\int \\frac{d}{dx}(\\mu(x)y)dx& =\\hfill & \\int \\mu(x)q(x)dx\\hfill \\ \\hfill \\mu(x)y& =\\hfill & \\int \\mu(x)q(x)dx\\hfill \\end{array}[\/latex]<\/p>\r\n

Divide both sides by [latex]\\mu(x)[\/latex]:<\/p>\r\n

[latex]y=\\frac{1}{\\mu(x)}\\left[\\int \\mu(x)q(x)dx+C\\right][\/latex]<\/p>\r\n

Since [latex]\\mu(x)[\/latex] was previously calculated, we are now finished.<\/p>\r\n\r\n

\r\n

integrating factor<\/h3>\r\nAn integrating factor is a function [latex]\\mu(x)[\/latex] that, when multiplied by a first-order linear differential equation, makes the left side equal to the derivative of a product. For the standard form [latex]y' + p(x)y = q(x)[\/latex], the integrating factor is:\r\n

[latex]\\mu(x) = e^{\\int p(x)dx}[\/latex]<\/p>\r\n\r\n<\/section>

When finding the integrating factor, we can set the constant of integration to zero since we only need one specific integrating factor. However, when solving for [latex]y[\/latex], we must include the constant [latex]C[\/latex] to get the general family of solutions.<\/section>
\r\n

Problem-Solving Strategy: Solving a First-order Linear Differential Equation<\/strong><\/p>\r\n\r\n

    \r\n \t
  1. Put the equation into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].<\/li>\r\n \t
  2. Calculate the integrating factor [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int p\\left(x\\right)dx}[\/latex].<\/li>\r\n \t
  3. Multiply both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex].<\/li>\r\n \t
  4. Integrate both sides of the equation obtained in step [latex]3[\/latex], and divide both sides by [latex]\\mu \\left(x\\right)[\/latex].<\/li>\r\n \t
  5. If there is an initial condition, determine the value of [latex]C[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    \r\n
    \r\n

    Find a general solution for the differential equation [latex]xy^{\\prime} +3y=4{x}^{2}-3x[\/latex]. Assume [latex]x>0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n

    \r\n
      \r\n \t
    1. To put this differential equation into standard form, divide both sides by [latex]x\\text{:}[\/latex] \r\n<\/span>\r\n
      [latex]y^{\\prime} +\\frac{3}{x}y=4x - 3[\/latex].<\/div>\r\n\r\n<\/span>\r\nTherefore [latex]p\\left(x\\right)=\\frac{3}{x}[\/latex] and [latex]q\\left(x\\right)=4x - 3[\/latex].<\/li>\r\n \t
    2. The integrating factor is [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int \\left(\\frac{3}{x}\\right)dx}={e}^{3\\text{ln}x}={x}^{3}[\/latex].<\/li>\r\n \t
    3. Multiplying both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex] gives us\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill {x}^{3}{y}^{\\prime }+{x}^{3}\\left(\\frac{3}{x}\\right)y& =\\hfill & {x}^{3}\\left(4x - 3\\right)\\hfill \\\\ \\hfill {x}^{3}{y}^{\\prime }+3{x}^{2}y& =\\hfill & 4{x}^{4}-3{x}^{3}\\hfill \\\\ \\hfill \\frac{d}{dx}\\left({x}^{3}y\\right)& =\\hfill & 4{x}^{4}-3{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n <\/li>\r\n \t
    4. Integrate both sides of the equation.\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left({x}^{3}y\\right)dx}& =\\hfill & {\\displaystyle\\int 4{x}^{4}-3{x}^{3}dx}\\hfill \\\\ \\hfill {x}^{3}y& =\\hfill & \\frac{4{x}^{5}}{5}-\\frac{3{x}^{4}}{4}+C\\hfill \\\\ \\hfill y& =\\hfill & \\frac{4{x}^{2}}{5}-\\frac{3x}{4}+C{x}^{-3}.\\hfill \\end{array}[\/latex]<\/div>\r\n <\/li>\r\n \t
    5. There is no initial value, so the problem is complete.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n
      Analysis<\/strong><\/div>\r\n

      You may have noticed the condition that was imposed on the differential equation; namely, [latex]x>0[\/latex]. For any nonzero value of [latex]C[\/latex], the general solution is not defined at [latex]x=0[\/latex]. Furthermore, when [latex]x<0[\/latex], the integrating factor changes. The integrating factor is given by [latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex]\u00a0as [latex]f\\left(x\\right)={e}^{\\displaystyle\\int p\\left(x\\right)dx}[\/latex]. For this [latex]p\\left(x\\right)[\/latex] we get<\/p>\r\n\r\n

      [latex]{e}^{\\displaystyle\\int p\\left(x\\right)dx=}{e}^{\\displaystyle\\int \\left(\\frac{3}{x}\\right)dx}={e}^{3\\text{ln}|x|}={|x|}^{3}[\/latex],<\/div>\r\n \r\n

      since [latex]x<0[\/latex]. The behavior of the general solution changes at [latex]x=0[\/latex] largely due to the fact that [latex]p\\left(x\\right)[\/latex] is not defined there.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

      Watch the following video to see the worked solution to the example above.