{"id":1562,"date":"2025-10-17T17:14:36","date_gmt":"2025-10-17T17:14:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1562"},"modified":"2025-11-11T15:14:15","modified_gmt":"2025-11-11T15:14:15","slug":"nonhomogeneous-linear-equations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/nonhomogeneous-linear-equations-learn-it-2\/","title":{"raw":"Nonhomogeneous Linear Equations: Learn It 2","rendered":"Nonhomogeneous Linear Equations: Learn It 2"},"content":{"raw":"

Undetermined Coefficients<\/h2>\r\nThe method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of [latex]r(x)[\/latex]. When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when [latex]r(x)[\/latex] has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form.\r\n\r\n
Find the general solution to [latex]y^{\\prime\\prime}+4y^\\prime+3y=3x[\/latex].[reveal-answer q=\"204756223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204756223\"]\r\n

The complementary equation is [latex]y^{\\prime\\prime}+4y^\\prime+3y=0[\/latex], with general solution [latex]c_1e^{-x}+c_2e^{-3x}[\/latex]. Since [latex]r(x)=3x[\/latex], the particular solution might have the form [latex]y_p(x)=Ax+B[\/latex]. If this is the case, then we have [latex]y_p^\\prime(x)=A[\/latex] and [latex]y_py^{\\prime\\prime}(x)=0[\/latex]. For [latex]y_p[\/latex] to be a solution to the differential equation, we must find values for [latex]A[\/latex] and [latex]B[\/latex] such that<\/p>\r\n

[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+4^\\prime+3y&=3x \\\\\r\n0+4(A)+3(Ax+B)&=3x \\\\\r\n3Ax+(4A+3B)&=3x\r\n\\end{aligned}[\/latex].<\/p>\r\n

Setting coefficients of like terms equal, we have<\/p>\r\n

[latex]\r\n\\begin{aligned}\r\n3A&=3 \\\\\r\n4A+3b&=0\r\n\\end{aligned}\r\n[\/latex].<\/p>\r\n

Then, [latex]A=1[\/latex] and [latex]B=-\\frac{4}{3}[\/latex], so [latex]y_p(x)=x-\\frac{4}{3}[\/latex] and the general solution is<\/p>\r\n

[latex]y(x)=c_1e^{-x}+c_2e^{-3x}+x-\\frac43[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

In the example above, notice that even though [latex]r(x)[\/latex] did not include a constant term, it was necessary to include the constant term in our guess. If we had assumed a solution of the form [latex]y_p=Ax[\/latex] (with no constant term), we would not have been able to find a solution.\r\n

If the function [latex]r(x)[\/latex] is a polynomial, our guess for the particular solution should be a polynomial of the same degree, and it must include all lower-order terms, regardless of whether they are present in [latex]r(x)[\/latex].<\/p>\r\n\r\n<\/section>

Find the general solution to [latex]y^{\\prime\\prime}-y^\\prime-2y=2e^{3x}[\/latex].[reveal-answer q=\"083272901\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"083272901\"]\r\n

The complementary equation is [latex]y^{\\prime\\prime}-y^\\prime-2y=0[\/latex], with the general solution [latex]c_1e^{-x}+c_2e^{2x}[\/latex]. Since [latex]r(x)=2e^{3x}[\/latex], the particular solution might have the form [latex]y_p=Ae^{3x}[\/latex]. Then, we have [latex]y_p^\\prime(x)=3Ae^{3x}[\/latex] and [latex]y^{\\prime\\prime}=9Ae^{3x}[\/latex]. For [latex]y_p[\/latex] to be a solution to the differential equation, we must find a value for [latex]A[\/latex] such that<\/p>\r\n

[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-y^\\prime-2y&=2e^{3x} \\\\\r\n9Ae^{3x}-3Ae^{3x}-2Ae^{3x}&=2Ae^{3x} \\\\\r\n4Ae^{3x}&=2Ae^{3x}\r\n\\end{aligned}[\/latex].<\/p>\r\n

So, [latex]4A=2[\/latex] and [latex]A=\\frac{1}{2}[\/latex]. Then, [latex]y_p(x)=\\left(\\frac12\\right)e^{3x}[\/latex], and the general solution is<\/p>\r\n

[latex]y(x)=c_1e^{-x}+c_2e^{2x}+\\frac12e^{3x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

When [latex]r(x)[\/latex] includes both sine and cosine terms, both terms must be present in the guess\u2014even if [latex]r(x)[\/latex] included only a sine term or only a cosine term.<\/section>The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of [latex]r(x)[\/latex] and the associated guesses for [latex]y_p(x)[\/latex] are summarized below.\r\n\r\n
\r\n

key forms for the method of undetermined coefficients<\/h3>\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]r(x)[\/latex]<\/th>\r\nInitial guess for [latex]y_p(x)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]k[\/latex] (a constant)<\/td>\r\n[latex]A[\/latex] (a constant)<\/td>\r\n<\/tr>\r\n
[latex]ax+b[\/latex]<\/td>\r\n[latex]Ax+B[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]ax^{2}+bx+c[\/latex]<\/td>\r\n[latex]Ax^{2}+Bx+C[\/latex]<\/td>\r\n<\/tr>\r\n
Higher-order polynomials<\/td>\r\nPolynomial of the same order as [latex]r(x)[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]ae^{\\lambda x}[\/latex]<\/td>\r\n[latex]Ae^{\\lambda x}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]a\\cos\\beta x+b\\sin\\beta x[\/latex]<\/td>\r\n[latex]A\\cos\\beta x+B\\sin\\beta x[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]ae^{\\alpha x}\\cos\\beta x+be^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n[latex]Ae^{\\alpha x}\\cos\\beta x+Be^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n<\/tr>\r\n
[latex](ax^2+bx+c)e^{\\lambda x}[\/latex]<\/td>\r\n[latex](Ax^2+Bx+C)e^{\\lambda x}[\/latex]<\/td>\r\n<\/tr>\r\n
[latex](a_2x^2+a_1x+a_0)\\cos\\beta x+(b_2x^2+b_1x+b_0)\\sin\\beta x[\/latex]<\/td>\r\n[latex](A_2x^2+A_1x+A_0)\\cos\\beta x+(B_2x^2+B_1x+B_0)\\sin\\beta x[\/latex]<\/td>\r\n<\/tr>\r\n
[latex](a_2x^2+a_1x+a_0)e^{\\alpha x}\\cos\\beta x+(b_2x^2+b_1x+b_0)e^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n[latex](A_2x^2+A_1x+A_0)e^{\\alpha x}\\cos\\beta x+(B_2x^2+B_1x+B_0)e^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Table 7.2\u00a0Key Forms for the Method of Undetermined Coefficients<\/span><\/p>\r\nNote: For polynomial guesses, the guess must include all lower-order terms, even if they have zero coefficients in [latex]r(x)[\/latex]. For trigonometric guesses, both sine and cosine terms must be included, even if only one appears in [latex]r(x)[\/latex].<\/em>\r\n\r\n<\/section>\r\n

A Key Pitfall<\/h3>\r\n

There is an important issue to watch for with this method. Consider the differential equation [latex]y^{\\prime\\prime}+5y^\\prime+6y=3e^{-2x}[\/latex]. Based on the form of [latex]r(x)[\/latex], we guess a particular solution of the form [latex]y_p(x)=Ae^{-2x}[\/latex]. But when we substitute this expression into the differential equation to find a value for [latex]A[\/latex], we run into a problem.<\/p>\r\n

We have [latex]y_p^\\prime(x)=-2Ae^{-2x}[\/latex] and [latex]y_p^{\\prime\\prime}(x)=4Ae^{-2x}[\/latex], so we want:<\/p>\r\n

[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+5y^\\prime+6y&=3e^{-2x} \\\\\r\n4Ae^{-2x}+\\left(-2Ae^{-2x}\\right)+6Ae^{-2x}&=3e^{-2x} \\\\\r\n4Ae^{-2x}-10Ae^{-2x}+6Ae^{-2x}&=3e^{-2x} \\\\\r\n0&=3e^{-2x}\r\n\\end{aligned}[\/latex],<\/p>\r\nwhich is not possible.\r\n

Looking closely, we see that the general solution to the complementary equation is [latex]c_1e^{-2x}+c_2e^{-3x}[\/latex]. The exponential function in [latex]r(x)[\/latex] is actually a solution to the complementary equation, so all the terms on the left side of the equation cancel out.<\/p>\r\n

We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by [latex]x[\/latex]. Using the new guess, [latex]y_p(x)=Axe^{-2x}[\/latex], we have:<\/p>\r\n

[latex]y_p^\\prime(x)=A\\left(e^{-2x}-2xe^{-2x}\\right)[\/latex]<\/p>\r\n

and<\/p>\r\n

[latex]y_p^{\\prime\\prime}(x)=-4Ae^{-2x}+4Axe^{-2x}[\/latex].<\/p>\r\n

Substitution gives<\/p>\r\n

[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+5y^\\prime+6y&=3e^{-2x} \\\\\r\n\\left(-4Ae^{-2x}+4Axe^{-2x}\\right)+5\\left(Ae^{-2x}-2Axe^{-2x}\\right)+6Axe^{-2x}&=3e^{-2x} \\\\\r\n-4Ae^{-2x}+4Axe^{-2x}-10Axe^{-2x}+6Axe^{-2x}&=3e^{-2x} \\\\\r\nAe^{-2x}&=3e^{-2x}\r\n\\end{aligned}[\/latex],<\/p>\r\nSo, [latex]A=3[\/latex] and [latex]y_p(x)=3xe^{-2x}[\/latex]. This gives us the following general solution:\r\n

[latex]\\large{y(x)=c_1e^{-2x}+c_2e^{-2x}+3xe^{-2x}}[\/latex].<\/p>\r\n\r\n

If [latex]xe^{-2x}[\/latex] were also a solution to the complementary equation, we would have to multiply by [latex]x[\/latex] again, and we would try [latex]y_p(x)=Ax^2e^{-2x}[\/latex].<\/section>
How to: Use the Method of Undetermined Coefficients<\/strong>\r\n
    \r\n \t
  1. Solve the complementary equation<\/strong>: Write down the general solution.<\/li>\r\n \t
  2. Make an initial guess for [latex]y_p(x)[\/latex]<\/strong>: Base your guess on the form of [latex]r(x)[\/latex].<\/li>\r\n \t
  3. Check for conflicts with the complementary solution<\/strong>: If any term in the guess for [latex]y_p(x)[\/latex] is a solution to the complementary equation, multiply the guess by [latex]x[\/latex]. Repeat this step until there are no terms in [latex]y_p(x)[\/latex] that solve the complementary equation.<\/li>\r\n \t
  4. Substitute and solve for coefficients<\/strong>: Substitute [latex]y_p(x)[\/latex] into the differential equation and equate like terms to find values for the unknown coefficients in [latex]y_p(x)[\/latex].<\/li>\r\n \t
  5. Write the general solution<\/strong>: Add the general solution to the complementary equation and the particular solution you just found.<\/li>\r\n<\/ol>\r\n<\/section>
    \r\n

    Find the general solutions to the following differential equations.<\/p>\r\n\r\n

      \r\n \t
    1. [latex]y^{\\prime\\prime}-9y=-6\\cos 3x[\/latex]<\/li>\r\n \t
    2. [latex]x^{\\prime\\prime}+2x'+x=4e^{-t}[\/latex]<\/li>\r\n \t
    3. [latex]y^{\\prime\\prime}-2y^\\prime+5y=10^{2}-3x-3[\/latex]<\/li>\r\n \t
    4. [latex]y^{\\prime\\prime}-3y^\\prime=-12t[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"694379198\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"694379198\"]\r\n
        \r\n \t
      1. The complementary equation is [latex]y^{\\prime\\prime}-9y=0[\/latex], which has the general solution [latex]c_1e^{3x}+c_2e^{-3x}[\/latex] (step 1). Based on the form of [latex]r(x)=-6\\cos3x[\/latex], our initial guess for the particular solution is [latex]y_p(x)=A\\cos3x+B\\sin3x[\/latex] (step 2). None of the terms in [latex]y_p(x)[\/latex] solve the complementary equation, so this is a valid guess (step 3).\r\n<\/span>Now we want to find values for [latex]A[\/latex] and [latex]B[\/latex], so substitute [latex]y_p[\/latex] into the differential equation. We have\r\n<\/span>\r\n

        [latex]\\large{y_p^\\prime(x)=-3A\\sin3x+3B\\cos3x\\text{ and }y_p^{\\prime\\prime}(x)=-9A\\cos3x-9B\\sin3x}[\/latex],<\/p>\r\n\r\n<\/span>so we want to find values of [latex]A[\/latex] and [latex]B[\/latex] such that\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\ny^{\\prime\\prime}-9y&=-6\\cos 3x \\\\\r\n-9A\\cos3x-9B\\sin3x-9(A\\cos3x+B\\sin3x)&=-6\\cos 3x \\\\\r\n-18A\\cos3x-18B\\sin3x&=-6\\cos 3x\r\n\\end{aligned}[\/latex]<\/p>\r\nTherefore,\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\n-18A&=-6 \\\\\r\n-18B&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<\/span>This gives [latex]A=\\frac{1}{3}[\/latex] and [latex]B=0[\/latex], so [latex]y_p(x)=\\left(\\frac13\\right)\\cos3x[\/latex] (step 4).\r\n<\/span>Putting everything together, we have the general solution\r\n<\/span>\r\n\r\n

        [latex]\\large{y(x)=c_1e^{3x}+c_2e^{-3x}+\\frac13\\cos3x}[\/latex].<\/center><\/li>\r\n \t
      2. The complementary equation is [latex]x^{\\prime\\prime}+2x^\\prime+x=0[\/latex], which has the general solution [latex]c_1e^{-t}+c_2te^{-t}[\/latex] (step 1). Based on the form [latex]r(t)=4e^{-t}[\/latex], our initial guess for the particular solution is [latex]x_p(t)=Ae^{-t}[\/latex] (step 2). However, we see that this guess solves the complementary equation, so we must multiply by [latex]t[\/latex], which gives a new guess: [latex]x_p(t)=Ate^{-t}[\/latex] (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by [latex]t[\/latex] again, which gives [latex]x_p(t)=At^2e^{-t}[\/latex] (step 3 again). Now, checking this guess, we see that [latex]x_p(t)[\/latex] does not solve the complementary equation, so this is a valid guess (step 3 yet again).\r\n<\/span>We now want to find a value for [latex]A[\/latex], so we substitute [latex]x_p[\/latex] into the differential equation. We have\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\nx_p(t)&=At^2e^{-t} \\\\\r\nx_p^\\prime(t)&=2Ate^{-t} -At^2e^{-t}\r\n\\end{aligned}[\/latex]<\/p>\r\nand [latex]x_p^{\\prime\\prime}=2Ae^{-t}-2Ate^{-t}-\\left(2Ate^{-t}-At^2e^{-t}\\right)=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}[\/latex].\r\nSubstituting into the differential equation, we want to find a value of [latex]A[\/latex] so that\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\nx^{\\prime\\prime}+2x^\\prime+x&=4e^{-t} \\\\\r\n2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+2(2Ate^{-t}-At^2e^{-t})+At^2e^{-t}&=4e^{-t} \\\\\r\n2Ae^{-t}=4e^{-t}\\end{aligned}[\/latex].<\/p>\r\n\r\nThis gives [latex]A=2[\/latex], so [latex]x_p(t)=2t^2e^{-t}[\/latex] (step 4). Putting everything together, we have the general solution\r\n<\/span>\r\n\r\n

        [latex]\\large{x(t)=c_1e^{-t}+c_2te^{-t}+2t^2e^{-t}}[\/latex].<\/center><\/li>\r\n \t
      3. The complementary equation is [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex], which has the general solution [latex]c_1e^x\\cos2x+c_2e^x\\sin2x[\/latex] (step 1). Based on the form [latex]r(x)=10x^2-3x-3[\/latex], our initial guess for the particular solution is [latex]y_p(x)=Ax^2+Bx+C[\/latex] (step 2). None of the terms in [latex]y_p(x)[\/latex] solve the complementary equation, so this is a valid guess (step 3). We now want to find values for [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], so we substitute [latex]y_p[\/latex] into the differential equation. We have [latex]y_p^\\prime(x)=2Ax+B[\/latex] and [latex]y_p^{\\prime\\prime}(x)=2A[\/latex], so we want to find values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] such that\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\ny^{\\prime\\prime}-2y^\\prime+5y&=10x^2-3x-3 \\\\\r\n2A-2(2Ax+b)+5(Ax^2+Bx+C)&=10x^2-3x-3 \\\\\r\n5Ax^2+(5B-4A)x+(5C-2B+2A)&=10x^2-3x-3\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore,\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\n5A&=10 \\\\\r\n5B-4A&=-3 \\\\\r\n5C-2B+2A&=-3\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<\/span>This gives [latex]A=2[\/latex], [latex]B=1[\/latex], and [latex]C=-1[\/latex], so [latex]y_p(x)=2x^2+x-1[\/latex] (step 4). Putting everything together, we have the general solution\r\n<\/span>\r\n\r\n

        [latex]y(x)=c_1e^x\\cos2x+c_2e^x\\sin2x+2x^2+x-1[\/latex].<\/center><\/li>\r\n \t
      4. The complementary equation is [latex]y^{\\prime\\prime}-3y^\\prime=0[\/latex], which has the general solution [latex]c_1e^{3t}+c_2[\/latex] (step 1). Based on the form [latex]r(t)=-12t[\/latex], our initial guess for the particular solution is [latex]y_p(t)=At+B[\/latex] (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by [latex]t[\/latex], which gives a new guess: [latex]y_P(t)=At^2+Bt[\/latex] (step 3). Checking this new guess, we see that none of the terms in [latex]y_p(t)[\/latex] solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for [latex]A[\/latex] and [latex]B[\/latex], so we substitute [latex]y_p[\/latex] into the differential equation. We have [latex]y_p^\\prime(t)=2At+B[\/latex] and [latex]y_p^{\\prime\\prime}(t)=2A[\/latex], so we want to find values of [latex]A[\/latex] and [latex]B[\/latex] such that\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\ny^{\\prime\\prime}-3y^\\prime&=-12t \\\\\r\n2A-3(2At+B)=-12t \\\\\r\n-6At+(2A-3B)&=-12t\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<\/span>Therefore,\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\n-6A&=-12 \\\\\r\n2A-3B&=0\r\n\\end{aligned}[\/latex].<\/p>\r\nThis gives [latex]A=2[\/latex] and [latex]b=4\/3[\/latex], so [latex]y_p(t)=2t^2+(4\/3)t[\/latex] (step 4). Putting everything together, we have the general solution\r\n\r\n

        [latex]y(t)=c_1e^{3t}+c_2+2t^2+\\frac43t[\/latex].<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

        Undetermined Coefficients<\/h2>\n

        The method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of [latex]r(x)[\/latex]. When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when [latex]r(x)[\/latex] has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form.<\/p>\n

        Find the general solution to [latex]y^{\\prime\\prime}+4y^\\prime+3y=3x[\/latex].<\/p>\n