{"id":834,"date":"2025-06-20T17:16:32","date_gmt":"2025-06-20T17:16:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=834"},"modified":"2025-09-10T15:18:28","modified_gmt":"2025-09-10T15:18:28","slug":"first-order-linear-equations-and-applications-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-learn-it-2\/","title":{"raw":"First-Order Linear Equations and Applications: Learn It 2","rendered":"First-Order Linear Equations and Applications: Learn It 2"},"content":{"raw":"

Standard Form<\/h2>\r\n

Consider the differential equation<\/p>\r\n\r\n

[latex]\\left(3{x}^{2}-4\\right){y}^{\\prime }+\\left(x - 3\\right)y=\\sin{x}[\/latex].<\/div>\r\n \r\n

Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of [latex]{y}^{\\prime }[\/latex] be equal to [latex]1[\/latex]. To make this happen, we divide both sides by [latex]3{x}^{2}-4[\/latex].<\/p>\r\n\r\n

[latex]{y}^{\\prime }+\\left(\\frac{x - 3}{3{x}^{2}-4}\\right)y=\\frac{\\sin{x}}{3{x}^{2}-4}[\/latex]<\/div>\r\n \r\n

This is called the standard form <\/strong>of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation.<\/p>\r\nReturning to our general definition, we can divide both sides of the equation by [latex]a\\left(x\\right)[\/latex]. This leads to the equation\r\n

[latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex].<\/div>\r\n

Now define [latex]p\\left(x\\right)=\\frac{b\\left(x\\right)}{a\\left(x\\right)}[\/latex] and [latex]q\\left(x\\right)=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. Then the definition becomes<\/p>\r\n\r\n

[latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex].<\/div>\r\n
We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.<\/div>\r\n
\r\n

standard form for first-order linear differential equations<\/h3>\r\n

The standard form of a first-order linear differential equation is:<\/p>\r\n

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

where [latex]p(x)[\/latex] and [latex]q(x)[\/latex] are functions of [latex]x[\/latex] only.<\/p>\r\n\r\n<\/section><\/div>\r\n

\r\n
\r\n

Put each of the following first-order linear differential equations into standard form. Identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for each equation.<\/p>\r\n\r\n

    \r\n \t
  1. [latex]y^{\\prime} =3x - 4y[\/latex]<\/li>\r\n \t
  2. [latex]\\frac{3xy^{\\prime} }{4y - 3}=2[\/latex] (here [latex]x>0[\/latex])<\/li>\r\n \t
  3. [latex]y=3y^{\\prime} -4{x}^{2}+5[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n
    \r\n
      \r\n \t
    1. Add [latex]4y[\/latex] to both sides:\r\n<\/span>\r\n
      [latex]y^{\\prime} +4y=3x[\/latex].<\/div>\r\n\r\n<\/span>\r\nIn this equation, [latex]p\\left(x\\right)=4[\/latex] and [latex]q\\left(x\\right)=3x[\/latex].<\/li>\r\n \t
    2. Multiply both sides by [latex]4y - 3[\/latex], then subtract [latex]8y[\/latex] from each side:\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill \\frac{3xy^{\\prime} }{4y - 3}& =\\hfill & 2\\hfill \\\\ \\hfill 3xy^{\\prime} & =\\hfill & 2\\left(4y - 3\\right)\\hfill \\\\ \\hfill 3xy^{\\prime} & =\\hfill & 8y - 6\\hfill \\\\ \\hfill 3xy^{\\prime} -8y& =\\hfill & -6.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nFinally, divide both sides by [latex]3x[\/latex] to make the coefficient of [latex]y^{\\prime} [\/latex] equal to [latex]1\\text{:}[\/latex] \r\n<\/span>\r\n
      [latex]y^{\\prime} -\\frac{8}{3x}y=-\\frac{2}{3x}[\/latex].<\/div>\r\nThis is allowable because in the original statement of this problem we assumed that [latex]x>0[\/latex]. (If [latex]x=0[\/latex] then the original equation becomes [latex]0=2[\/latex], which is clearly a false statement.)\r\n<\/span>\r\nIn this equation, [latex]p\\left(x\\right)=-\\frac{8}{3x}[\/latex] and [latex]q\\left(x\\right)=-\\frac{2}{3x}[\/latex].<\/li>\r\n \t
    3. Subtract [latex]y[\/latex] from each side and add [latex]4{x}^{2}-5\\text{:}[\/latex] \r\n<\/span>\r\n
      [latex]3y^{\\prime} -y=4{x}^{2}-5[\/latex].<\/div>\r\n\r\n<\/span>\r\nNext divide both sides by [latex]3\\text{:}[\/latex] \r\n<\/span>\r\n
      [latex]y^{\\prime} -\\frac{1}{3}y=\\frac{4}{3}{x}^{2}-\\frac{5}{3}[\/latex].<\/div>\r\n\r\n<\/span>\r\nIn this equation, [latex]p\\left(x\\right)=-\\frac{1}{3}[\/latex] and [latex]q\\left(x\\right)=\\frac{4}{3}{x}^{2}-\\frac{5}{3}[\/latex].<\/li>\r\n<\/ol>\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.