{"id":1547,"date":"2025-10-17T17:01:29","date_gmt":"2025-10-17T17:01:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1547"},"modified":"2025-11-11T14:30:56","modified_gmt":"2025-11-11T14:30:56","slug":"linear-homogeneous-equations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/linear-homogeneous-equations-learn-it-2\/","title":{"raw":"Linear Homogeneous Equations: Learn It 2","rendered":"Linear Homogeneous Equations: Learn It 2"},"content":{"raw":"

Finding the General Solution<\/h2>\r\n

Although finding any solution to a differential equation is important, we often want to find all solutions to a differential equation. We call this the general solution<\/strong>.<\/p>\r\n\r\n

Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation.<\/section>An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact in the following theorem.\r\n\r\n
\r\n

superposition principle<\/h3>\r\n

If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are solutions to a linear homogeneous differential equation, then the function<\/p>\r\n

[latex]y(x)=c_1y_1(x)+c_2y_2(x)[\/latex],<\/p>\r\nwhere [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants, is also a solution.\r\n\r\n<\/section>

\r\n

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

[latex]\\large{y^{\\prime\\prime}-4y^\\prime-5y=0}[\/latex].<\/p>\r\n

Given that [latex]e^{-x}[\/latex] and [latex]e^{5x}[\/latex] are solutions to this differential equation, show that [latex]4e^{-x}+e^{5x}[\/latex] is a solution.<\/p>\r\n[reveal-answer q=\"830429349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830429349\"]\r\n

We have<\/p>\r\n

[latex]y(x)=4e^{-x}+e^{5x},\\text{ so }y^\\prime(x)=-4e^{-x}+5e^{5x}\\text{ and }y^{\\prime\\prime}=4e^{-x}+25e^{5x}[\/latex].<\/p>\r\n

Then<\/p>\r\n

[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-4y^\\prime-5y&=\\left(4e^{-x}+25e^{5x}\\right)-4\\left(-4e^{-x}+5e^{5x}\\right)-5\\left(4e^{-x}+e^{5x}\\right) \\\\\r\n&=4e^{-x}+25e^{5x}+16e^{-x}-20e^{5x}-20e^{-x}-5e^{5x} \\\\\r\n&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n

Thus, [latex]y(x)=4e^{-x}+e^{5x}[\/latex] is a solution.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation:<\/p>\r\n

[latex]x''+7x'+12x=0[\/latex].<\/p>\r\n

Both [latex]e^{-3t}[\/latex] and [latex]2e^{-3t}[\/latex] are solutions (check this). However, [latex]x(t)=c_1e^{-3y}+c_2\\left(2e^{-3t}\\right)[\/latex] is\u00a0not<\/em>\u00a0the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function [latex]e^{-4t}[\/latex], which is also a solution to the differential equation.<\/p>\r\n

To find the general solution to a second-order differential equation, we must find two linearly independent<\/strong> solutions.<\/p>\r\n\r\n

\r\n

linearly independent<\/h3>\r\nA set of functions [latex]f_1(x),f_2(x),\\ldots,f_n(x)[\/latex] is said to be linearly dependent<\/span><\/strong>\u00a0if there are constants [latex]c_1,c_2,\\ldots c_n[\/latex], not all zero, such that [latex]c_1f_1(x)+c_2f_2(x)+\\cdots c_nf_n(x)=0[\/latex] for all [latex]x[\/latex] over the interval of interest.\r\n\r\n \r\n\r\nA set of functions that is not linearly dependent is said to be\u00a0linearly independent<\/span>.<\/strong>\r\n\r\n<\/section>\r\n
\r\n

You will usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.<\/p>\r\n

If the functions meet these conditions, then they are linearly dependent. If one of the functions is identically zero\u2014say, [latex]f_2(x)\\equiv0[\/latex]\u2014then choose [latex]c_1=0[\/latex] and [latex]c_2=1[\/latex], and the condition for linear dependence is satisfied. If, on the other hand, neither [latex]f_1(x)[\/latex] nor [latex]f_2(x)[\/latex] is identically zero, but [latex]f_1(x)=Cf_2(x)[\/latex] for some constant [latex]C[\/latex], then choose [latex]c_1=\\frac{1}{C}[\/latex] and [latex]c_2=-1[\/latex], and again, the condition is satisfied.<\/p>\r\n

Conversely, if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume [latex]f_1(x)[\/latex] and [latex]f_2(x)[\/latex] are linearly dependent. Then, there are constants, [latex]c_1[\/latex] and [latex]c_2[\/latex], not both zero, such that:<\/p>\r\n

[latex]c_1f_1(x)+c_2f_2(x)=0[\/latex]<\/p>\r\n

for all [latex]x[\/latex] over the interval of interest. Then,<\/p>\r\n\r\n<\/div>\r\n

[latex]c_1f_1(x)=-c_2f_2(x)[\/latex].<\/p>\r\n\r\n

\r\n

Now, since we stated that [latex]c_1[\/latex] and [latex]c_2[\/latex] can\u2019t both be zero, assume [latex]c_2\\ne0[\/latex]. Then, there are two cases: either [latex]c_1=0[\/latex] or [latex]c_1\\ne0[\/latex]. If [latex]c_1=0[\/latex], then:<\/p>\r\n

[latex]\\begin{aligned}\r\n0&=-c_2f_2(x) \\\\\r\n0&=f_2(x)\r\n\\end{aligned}[\/latex],<\/p>\r\n

so one of the functions is identically zero. Now suppose [latex]c_1\\ne0[\/latex]. Then,<\/p>\r\n

[latex]f_1(x)=\\left(-\\frac{c_2}{c_1}\\right)f_2(x)[\/latex]<\/p>\r\n

and we see that the functions are constant multiples of one another.<\/p>\r\n\r\n

\r\n

linear dependence of two functions<\/h3>\r\nTwo functions, [latex]f_1(x)[\/latex] and [latex]f_2(x)[\/latex], are said to be linearly dependent<\/strong> if either one of them is identically zero or if [latex]f_1(x)=Cf_2(x)[\/latex] for some constant [latex]C[\/latex] and for all [latex]x[\/latex] over the interval of interest.\r\n\r\n \r\n\r\nFunctions that are not linearly dependent are said to be linearly independent<\/strong>.\r\n\r\n<\/section>
\r\n

Determine whether the following pairs of functions are linearly dependent or linearly independent.<\/p>\r\n\r\n

    \r\n \t
  1. [latex]f_1(x)=x^{2}[\/latex], [latex]f_2(x)=5x^{2}[\/latex]<\/li>\r\n \t
  2. [latex]f_1(x)=\\sin x[\/latex],\u00a0[latex]f_2(x)=\\cos x[\/latex]<\/li>\r\n \t
  3. [latex]f_1(x)=e^{3x}[\/latex],\u00a0[latex]f_2(x)=e^{-3x}[\/latex]<\/li>\r\n \t
  4. [latex]f_1(x)=3x[\/latex],\u00a0[latex]f_2(x)=3x+1[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"465788210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"465788210\"]\r\n
      \r\n \t
    1. [latex]f_2(x)=5f_1(x)[\/latex], so the functions are linearly dependent.<\/li>\r\n \t
    2. There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent.<\/li>\r\n \t
    3. There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent. Don\u2019t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.<\/li>\r\n \t
    4. There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution.\r\n\r\n
      \r\n

      general solution to a homogeneous equation<\/h3>\r\n

      If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by:<\/p>\r\n

      [latex]y(x)=c_1y_1(x)+c_2y_2(x)[\/latex],<\/p>\r\n

      where[latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n\r\n<\/section>When we say a family of functions is the general solution to a differential equation, we mean that (1) every expression of that form is a solution and (2) every solution to the differential equation can be written in that form. This makes this theorem extremely powerful. If we can find two linearly independent solutions to a differential equation, we have effectively found all solutions to the differential equation.\r\n\r\n

      If [latex]y_1(t)=e^{3t}[\/latex] and [latex]y_2(t)=e^{-3t}[\/latex] are solutions to [latex]y''-9y=0[\/latex], what is the general solution?[reveal-answer q=\"423749834\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"423749834\"]Note that [latex]y_1[\/latex] and\u00a0[latex]y_2[\/latex] are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is [latex]y(t)=c_1e^{3t}+c_2e^{-3t}[\/latex].[\/hidden-answer]<\/section><\/div>\r\n<\/div>","rendered":"

      Finding the General Solution<\/h2>\n

      Although finding any solution to a differential equation is important, we often want to find all solutions to a differential equation. We call this the general solution<\/strong>.<\/p>\n

      Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation.<\/section>\n

      An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact in the following theorem.<\/p>\n

      \n

      superposition principle<\/h3>\n

      If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are solutions to a linear homogeneous differential equation, then the function<\/p>\n

      [latex]y(x)=c_1y_1(x)+c_2y_2(x)[\/latex],<\/p>\n

      where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants, is also a solution.<\/p>\n<\/section>\n

      \n

      Consider the differential equation<\/p>\n

      [latex]\\large{y^{\\prime\\prime}-4y^\\prime-5y=0}[\/latex].<\/p>\n

      Given that [latex]e^{-x}[\/latex] and [latex]e^{5x}[\/latex] are solutions to this differential equation, show that [latex]4e^{-x}+e^{5x}[\/latex] is a solution.<\/p>\n