{"id":1563,"date":"2025-10-17T17:14:38","date_gmt":"2025-10-17T17:14:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1563"},"modified":"2025-11-11T15:20:59","modified_gmt":"2025-11-11T15:20:59","slug":"nonhomogeneous-linear-equations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/nonhomogeneous-linear-equations-learn-it-3\/","title":{"raw":"Nonhomogeneous Linear Equations: Learn It 3","rendered":"Nonhomogeneous Linear Equations: Learn It 3"},"content":{"raw":"

Variation of Parameters<\/h2>\r\n

Sometimes, [latex]r(x)[\/latex] is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the method of variation of parameters<\/strong>.<\/p>\r\n

To simplify our calculations, we are going to divide the differential equation through by [latex]a[\/latex], so we have a leading coefficient of [latex]1[\/latex]. Then the differential equation has the form:<\/p>\r\n\r\n

[latex]y^{\\prime\\prime}+py^\\prime+qy=r(x)[\/latex],<\/center>\r\n
<\/div>\r\nwhere [latex]p[\/latex] and [latex]q[\/latex] are constants.\r\n

If the general solution to the complementary equation is given by [latex]c_1y_1(x)+c_2y_2(x)[\/latex], we are going to look for a particular solution of the form [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex]. In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of [latex]x[\/latex], rather than constants. We want to find functions [latex]u(x)[\/latex] and [latex]v(x)[\/latex] such that [latex]y_p(x)[\/latex] satisfies the differential equation.<\/p>\r\n

We have:<\/p>\r\n\r\n<\/div>\r\n

[latex]\\begin{aligned}\r\ny_p&=uy_1+vy_2 \\\\\r\ny_p^\\prime&=u^\\prime+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime \\\\\r\ny_p^{\\prime\\prime}&=(u^\\prime y_1+v^\\prime y_2)^\\prime+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}\r\n\\end{aligned}[\/latex].<\/p>\r\n

Substituting into the differential equation, we obtain:<\/p>\r\n

[latex]\\begin{aligned}\r\ny_p^{\\prime\\prime}+py_p^\\prime+qy_p&=[(u^\\prime y_1+v^\\prime y_2)+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}]+p[u^\\prime y_1+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime]+q[uy_1+vy_2] \\\\\r\n&=u[y_1^{\\prime\\prime}+py_1^\\prime+qy_1]+v[y_2^{\\prime\\prime}+py_2^\\prime+qy_2]+(u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)\r\n\\end{aligned}[\/latex].<\/p>\r\n

Note that [latex]y_1[\/latex] and [latex]y_2[\/latex] are solutions to the complementary equation, so the first two terms are zero. Thus, we have:<\/p>\r\n

[latex](u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)=r(x)[\/latex].<\/p>\r\nIf we simplify this equation by imposing the additional condition [latex]u^\\prime y_1+v^\\prime y_2=0[\/latex], the first two terms are zero, and this reduces to [latex]u^\\prime y_1^\\prime+v^\\prime y_2^\\prime=r(x)[\/latex]. So, with this additional condition, we have a system of two equations in two unknowns:\r\n

[latex]\\begin{aligned}\r\nu^\\prime y_1+v^\\prime y_2&=0 \\\\\r\nu^\\prime y_1^\\prime+v^\\prime y_2^\\prime&=r(x)\r\n\\end{aligned}[\/latex].<\/p>\r\n

Solving this system gives us [latex]u^\\prime[\/latex] and [latex]v^\\prime[\/latex], which we can integrate to find [latex]u[\/latex] and [latex]v[\/latex]. Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the differential equation.<\/p>\r\n

Solving this system of equations is sometimes challenging, so let's review Cramer's rule, which allows us to solve the system of equations using determinants.<\/p>\r\n\r\n

\r\n

Cramer's rule<\/h3>\r\n

The system of equations:<\/p>\r\n

[latex]\\begin{aligned}\r\na_1z_1+b_1z_2=r_1 \\\\\r\na_2z_1+b_2z_2&=r_2\r\n\\end{aligned}[\/latex]<\/p>\r\n

has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by:<\/p>\r\n

[latex]z_1=\\frac{\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}[\/latex] and\u00a0[latex]z_2=\\frac{\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}[\/latex].<\/p>\r\n\r\n<\/section>

\r\n

Use Cramer\u2019s rule to solve the following system of equations.<\/p>\r\n

[latex]\\begin{aligned}\r\nx^2z_1+2xz_2&=0 \\\\\r\nz_1-3x^2z_2&=2x\r\n\\end{aligned}[\/latex].<\/p>\r\n[reveal-answer q=\"923921637\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923921637\"]\r\n

We have<\/p>\r\n

[latex]\\begin{aligned}\r\na_1(x)&=x^2 \\\\\r\na_2(x)&=1 \\\\\r\nb_1(x)&=2x \\\\\r\nb_2(x)&=-3x^2 \\\\\r\nr_1(x)&=0 \\\\\r\nr_2(x)&=2x\r\n\\end{aligned}[\/latex].<\/p>\r\n

Then,<\/p>\r\n

[latex]\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}=\\begin{vmatrix}x^2&2x \\\\ 1&-3x^2\\end{vmatrix}=-3x^4-2x[\/latex]<\/p>\r\n

and<\/p>\r\n

[latex]\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}=\\begin{vmatrix}0&2x \\\\ 2x&-3x^2\\end{vmatrix}=0-4x^2=-4x^2[\/latex].<\/p>\r\n

Thus,<\/p>\r\n

[latex]z_1=\\frac{\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}=\\frac{-4x^2}{-3x^4-2x}=\\frac{4x}{3x^3+2}[\/latex].<\/p>\r\n

In addition,<\/p>\r\n

[latex]\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}=\\begin{vmatrix}x^2&0 \\\\1&2x\\end{vmatrix}=2x^3-0=2x^3[\/latex].<\/p>\r\n

Thus,<\/p>\r\n

[latex]z_2=\\frac{\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}=\\frac{2x^3}{-3x^4-2x}=\\frac{-2x^2}{3x^3+2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

How to: Use the Method of Variation of Parameters<\/strong>\r\n
    \r\n \t
  1. Solve the complementary equation<\/strong>: Write down the general solution [latex]c_1y_1(x)+c_2y_2(x)[\/latex].<\/li>\r\n \t
  2. Find [latex]u^\\prime(x)[\/latex] and [latex]v^\\prime(x)[\/latex]<\/strong>: Use Cramer's rule or another suitable technique to find functions [latex]u^\\prime(x)[\/latex] and [latex]v^\\prime(x)[\/latex] satisfying [latex]\\begin{aligned} u^\\prime y_1+v^\\prime y_2&=0 \\\\ u^\\prime y_1^\\prime+v^\\prime y_2^\\prime&=r(x) \\end{aligned}[\/latex]<\/li>\r\n \t
  3. Integrate to find [latex]u[\/latex] and [latex]v[\/latex]<\/strong>: Integrate [latex]u^\\prime[\/latex] and [latex]v^\\prime[\/latex] to find [latex]u(x)[\/latex] and [latex]v(x)[\/latex]. Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the equation.<\/li>\r\n \t
  4. Form the particular solution<\/strong>: Calculate [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex].<\/li>\r\n \t
  5. Write the general solution<\/strong>: Add the general solution to the complementary equation and the particular solution found in step 4.<\/li>\r\n<\/ol>\r\n<\/section>
    \r\n

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

      \r\n \t
    1. [latex]y^{\\prime\\prime}-2y^\\prime+y=\\frac{e^t}{t^2}[\/latex]<\/li>\r\n \t
    2. [latex]y^{\\prime\\prime}+y=3\\sin^2x[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"982348083\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"982348083\"]\r\n
        \r\n \t
      1. The complementary equation is [latex]y^{\\prime\\prime}-2y^\\prime+y=0[\/latex] with associated general solution [latex]c_1e^t+c_2te^t[\/latex]. Therefore, [latex]y_1(t)=e^t[\/latex] and [latex]y_2(t)=te^t[\/latex]. Calculating the derivatives, we get [latex]y_1^\\prime(t)=e^t[\/latex] and [latex]y_2^\\prime(t)=e^t+te^t[\/latex] (step 1). Then, we want to find functions [latex]u'(t)[\/latex] and [latex]v'(t)[\/latex] so that\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\nu^\\prime e^t+v^\\prime te^t&=0 \\\\\r\nu^\\prime e^t+v^\\prime(e^t+te^t)&=\\frac{e^t}{t^2}\r\n\\end{aligned}\r\n[\/latex]<\/p>\r\n\r\n<\/span>Applying Cramer\u2019s rule, we have\r\n<\/span>\r\n

        [latex]u^\\prime=\\frac{\\begin{vmatrix}0&te^t \\\\ \\frac{e^t}{t^2}&e^t+te^t\\end{vmatrix}}{\\begin{vmatrix}e^t&te^t \\\\e^t&e^t+te^t\\end{vmatrix}}=\\frac{0-te^t\\left(\\frac{e^t}{t^2}\\right)}{e^t(e^t+te^t)-e^tte^t}=\\frac{-\\frac{e^{2t}}{t}}{e^{2t}}=-\\frac1t[\/latex]<\/p>\r\nand\r\n<\/span>\r\n

        \r\n

        [latex]v^\\prime=\\frac{\\begin{vmatrix}e^t&0\\\\e^t&\\frac{e^t}{t^2}\\end{vmatrix}}{\\begin{vmatrix}e^t&te^t\\\\e^t&e^t+te^t\\end{vmatrix}}=\\frac{e^t\\left(\\frac{e^t}{t^2}\\right)}{e^{2t}}=\\frac{1}{t^2}[\/latex] (step 2).<\/p>\r\n\r\n<\/div>\r\nIntegrating, we get\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\nu&=-\\displaystyle\\int\\frac1t \\ dt=-\\ln|t| \\\\\r\nv&=\\displaystyle\\int\\frac{1}{t^2}dt=-\\frac{1}{t}\\text{(step 3)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/span>Then we have\r\n<\/span>\r\n

        [latex]\\begin{aligned}\r\ny_p&=-e^t\\ln|t|-\\frac{1}{t}te^t \\\\\r\n&=-e^t\\ln|t|-e^t \\text{(step 4)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n<\/span>The [latex]e^{t}[\/latex] term is a solution to the complementary equation, so we don\u2019t need to carry that term into our general solution explicitly. The general solution is\r\n<\/span>\r\n

        [latex]y(t)=c_1e^t+c_2te^t-e^t\\ln|t|[\/latex] (step 5).<\/p>\r\n<\/li>\r\n \t

      2. The complementary equation is [latex]y^{\\prime\\prime}+y+0[\/latex] with associated general solution [latex]c_1\\cos x+c_2\\sin x[\/latex]. So, [latex]y_1(x)=\\cos x[\/latex] and [latex]y_2(x)=\\sin x[\/latex] (step 1). Then, we want to find functions [latex]u'(x)[\/latex] and [latex]v'(x)[\/latex] such that\r\n<\/span>\r\n

        [latex]\\begin{aligned} u^\\prime\\cos x+v^\\prime\\sin x&=0 \\\\ -y^\\prime\\sin x+v^\\prime\\cos x&=3\\sin^2x \\end{aligned}[\/latex].<\/p>\r\n\r\n<\/span>Applying Cramer\u2019s rule, we have\r\n<\/span>\r\n

        [latex]u^\\prime=\\frac{\\begin{vmatrix}0&\\sin x \\\\3\\sin^2x&\\cos x\\end{vmatrix}}{\\begin{vmatrix}\\cos x&\\sin x \\\\ -\\sin x&\\cos x\\end{vmatrix}}=\\frac{0-3\\sin^3x}{\\cos^2x+\\sin^2x}=-3\\sin^3x[\/latex]<\/p>\r\nand\r\n<\/span>\r\n

        [latex]v^\\prime=\\frac{\\begin{vmatrix}\\cos x&0\\\\ -\\sin x&3\\sin^2x\\end{vmatrix}}{\\begin{vmatrix}\\cos x&\\sin x \\\\-\\sin x&\\cos x\\end{vmatrix}}=\\frac{3\\sin^2x\\cos x}{1}=3\\sin^2x\\cos x[\/latex] (step 2).<\/p>\r\nIntegrating first to find [latex]u[\/latex], we get\r\n<\/span>\r\n

        [latex]u=\\displaystyle\\int-3\\sin^3xdx=-3\\left[-\\frac13\\sin^2x\\cos x+\\frac23\\displaystyle\\int\\sin xdx\\right]=\\sin^2x\\cos x+2\\cos x[\/latex] (step 3).<\/p>\r\n\r\n<\/span>Now, we integrate to find [latex]v[\/latex]. Using substitution (with [latex]w=\\sin x[\/latex]), we get\r\n<\/span>\r\n

        [latex]v=\\displaystyle\\int3\\sin^2x\\cos xdx=\\displaystyle\\int3w^2dw=w^3=\\sin^3x[\/latex].<\/p>\r\n\r\n<\/span>Then,\r\n<\/span>\r\n

        [latex]\r\n\\begin{aligned}\r\ny_p&=(\\sin^2x\\cos x+2\\cos x)\\cos x+(\\sin^3x)\\sin x \\\\\r\n&=\\sin^2x\\cos^2x+2\\cos^2x+\\sin^4x \\\\\r\n&=2\\cos^2x+\\sin^2x(\\cos^2x+\\sin^2x) \\ \\ \\ \\ (step 4). \\\\\r\n&=2\\cos^2x+\\sin^2 \\\\\r\n&=\\cos^2x+1\r\n\\end{aligned}\r\n[\/latex]<\/p>\r\nThe general solution is\r\n

        [latex]y(x)=c_1\\cos x+c_2\\sin x+1+\\cos^2x[\/latex] (step 5).<\/p>\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/section>","rendered":"

        Variation of Parameters<\/h2>\n

        Sometimes, [latex]r(x)[\/latex] is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the method of variation of parameters<\/strong>.<\/p>\n

        To simplify our calculations, we are going to divide the differential equation through by [latex]a[\/latex], so we have a leading coefficient of [latex]1[\/latex]. Then the differential equation has the form:<\/p>\n

        \n
        [latex]y^{\\prime\\prime}+py^\\prime+qy=r(x)[\/latex],<\/div>\n
        <\/div>\n

        where [latex]p[\/latex] and [latex]q[\/latex] are constants.<\/p>\n

        If the general solution to the complementary equation is given by [latex]c_1y_1(x)+c_2y_2(x)[\/latex], we are going to look for a particular solution of the form [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex]. In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of [latex]x[\/latex], rather than constants. We want to find functions [latex]u(x)[\/latex] and [latex]v(x)[\/latex] such that [latex]y_p(x)[\/latex] satisfies the differential equation.<\/p>\n

        We have:<\/p>\n<\/div>\n

        [latex]\\begin{aligned} y_p&=uy_1+vy_2 \\\\ y_p^\\prime&=u^\\prime+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime \\\\ y_p^{\\prime\\prime}&=(u^\\prime y_1+v^\\prime y_2)^\\prime+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime} \\end{aligned}[\/latex].<\/p>\n

        Substituting into the differential equation, we obtain:<\/p>\n

        [latex]\\begin{aligned} y_p^{\\prime\\prime}+py_p^\\prime+qy_p&=[(u^\\prime y_1+v^\\prime y_2)+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}]+p[u^\\prime y_1+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime]+q[uy_1+vy_2] \\\\ &=u[y_1^{\\prime\\prime}+py_1^\\prime+qy_1]+v[y_2^{\\prime\\prime}+py_2^\\prime+qy_2]+(u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime) \\end{aligned}[\/latex].<\/p>\n

        Note that [latex]y_1[\/latex] and [latex]y_2[\/latex] are solutions to the complementary equation, so the first two terms are zero. Thus, we have:<\/p>\n

        [latex](u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)=r(x)[\/latex].<\/p>\n

        If we simplify this equation by imposing the additional condition [latex]u^\\prime y_1+v^\\prime y_2=0[\/latex], the first two terms are zero, and this reduces to [latex]u^\\prime y_1^\\prime+v^\\prime y_2^\\prime=r(x)[\/latex]. So, with this additional condition, we have a system of two equations in two unknowns:<\/p>\n

        [latex]\\begin{aligned} u^\\prime y_1+v^\\prime y_2&=0 \\\\ u^\\prime y_1^\\prime+v^\\prime y_2^\\prime&=r(x) \\end{aligned}[\/latex].<\/p>\n

        Solving this system gives us [latex]u^\\prime[\/latex] and [latex]v^\\prime[\/latex], which we can integrate to find [latex]u[\/latex] and [latex]v[\/latex]. Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the differential equation.<\/p>\n

        Solving this system of equations is sometimes challenging, so let’s review Cramer’s rule, which allows us to solve the system of equations using determinants.<\/p>\n

        \n

        Cramer’s rule<\/h3>\n

        The system of equations:<\/p>\n

        [latex]\\begin{aligned} a_1z_1+b_1z_2=r_1 \\\\ a_2z_1+b_2z_2&=r_2 \\end{aligned}[\/latex]<\/p>\n

        has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by:<\/p>\n

        [latex]z_1=\\frac{\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}[\/latex] and\u00a0[latex]z_2=\\frac{\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}[\/latex].<\/p>\n<\/section>\n

        \n

        Use Cramer\u2019s rule to solve the following system of equations.<\/p>\n

        [latex]\\begin{aligned} x^2z_1+2xz_2&=0 \\\\ z_1-3x^2z_2&=2x \\end{aligned}[\/latex].<\/p>\n