17.2: Nonhomogeneous Linear Equations
( \newcommand{\kernel}{\mathrm{null}\,}\)
- Write the general solution to a nonhomogeneous differential equation.
- Solve a nonhomogeneous differential equation by the method of undetermined coefficients.
- Solve a nonhomogeneous differential equation by the method of variation of parameters.
In this section, we examine how to solve nonhomogeneous differential equations. The terminology and methods are different from those we used for homogeneous equations, so let’s start by defining some new terms.
General Solution to a Nonhomogeneous Linear Equation
Consider the nonhomogeneous linear differential equation
a2(x)y″+a1(x)y′+a0(x)y=r(x).
The associated homogeneous equation
a2(x)y″+a1(x)y′+a0(x)y=0
is called the complementary equation. We will see that solving the complementary equation is an important step in solving a nonhomogeneous differential equation.
A solution yp(x) of a differential equation that contains no arbitrary constants is called a particular solution to the equation.
Let yp(x) be any particular solution to the nonhomogeneous linear differential equation
a2(x)y″+a1(x)y′+a0(x)y=r(x).
Also, let c1y1(x)+c2y2(x) denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by
y(x)=c1y1(x)+c2y2(x)+yp(x).
To prove y(x) is the general solution, we must first show that it solves the differential equation and, second, that any solution to the differential equation can be written in that form. Substituting y(x) into the differential equation, we have
a2(x)y″+a1(x)y′+a0(x)y=a2(x)(c1y1+c2y2+yp)″+a1(x)(c1y1+c2y2+yp)′+a0(x)(c1y1+c2y2+yp)=[a2(x)(c1y1+c2y2)″+a1(x)(c1y1+c2y2)′+a0(x)(c1y1+c2y2)]+a2(x)yp″+a1(x)yp′+a0(x)yp=0+r(x)=r(x).
So y(x) is a solution.
Now, let z(x) be any solution to a2(x)y″+a1(x)y′+a0(x)y=r(x). Then
a2(x)(z−yp)″+a1(x)(z−yp)′+a0(x)(z−yp)=(a2(x)z″+a1(x)z′+a0(x)z)−(a2(x)yp″+a1(x)yp′+a0(x)yp)=r(x)−r(x)=0,
so z(x)−yp(x) is a solution to the complementary equation. But, c1y1(x)+c2y2(x) is the general solution to the complementary equation, so there are constants c1 and c2 such that
z(x)−yp(x)=c1y1(x)+c2y2(x).
Hence, we see that
z(x)=c1y1(x)+c2y2(x)+yp(x).
Given that yp(x)=x is a particular solution to the differential equation y″+y=x, write the general solution and check by verifying that the solution satisfies the equation.
Solution
The complementary equation is y″+y=0, which has the general solution c1cosx+c2sinx. So, the general solution to the nonhomogeneous equation is
y(x)=c1cosx+c2sinx+x.
To verify that this is a solution, substitute it into the differential equation. We have
y′(x)=−c1sinx+c2cosx+1
and
y″(x)=−c1cosx−c2sinx.
Then
y″(x)+y(x)=−c1cosx−c2sinx+c1cosx+c2sinx+x=x.
So, y(x) is a solution to y″+y=x.
Given that yp(x)=−2 is a particular solution to y″−3y′−4y=8, write the general solution and verify that the general solution satisfies the equation.
- Hint
-
Find the general solution to the complementary equation.
- Answer
-
y(x)=c1e−x+c2e4x−2
In the preceding section, we learned how to solve homogeneous equations with constant coefficients. Therefore, for nonhomogeneous equations of the form ay″+by′+cy=r(x), we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. We now examine two techniques for this: the method of undetermined coefficients and the method of variation of parameters.
Undetermined Coefficients
The method of undetermined coefficients involves making educated guesses about the form of the particular solution based on the form of r(x). When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when r(x) has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let’s look at some examples to see how this works.
Find the general solution to y″+4y′+3y=3x.
Solution
The complementary equation is y″+4y′+3y=0, with general solution c1e−x+c2e−3x. Since r(x)=3x, the particular solution might have the form yp(x)=Ax+B. If this is the case, then we have yp′(x)=A and yp″(x)=0. For yp to be a solution to the differential equation, we must find values for A and B such that
y″+4y′+3y=3x0+4(A)+3(Ax+B)=3x3Ax+(4A+3B)=3x.
Setting coefficients of like terms equal, we have
3A=34A+3B=0.
Then, A=1 and B=−43, so yp(x)=x−43 and the general solution is
y(x)=c1e−x+c2e−3x+x−43.
In Example 17.2.2, notice that even though r(x) did not include a constant term, it was necessary for us to include the constant term in our guess. If we had assumed a solution of the form yp=Ax (with no constant term), we would not have been able to find a solution. (Verify this!) If the function r(x) 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 r(x).
Find the general solution to y″−y′−2y=2e3x.
Solution
The complementary equation is y″−y′−2y=0, with the general solution c1e−x+c2e2x. Since r(x)=2e3x, the particular solution might have the form yp(x)=Ae3x. Then, we have yp′(x)=3Ae3x and yp″(x)=9Ae3x. For yp to be a solution to the differential equation, we must find a value for A such that
y″−y′−2y=2e3x9Ae3x−3Ae3x−2Ae3x=2e3x4Ae3x=2e3x.
So, 4A=2 and A=1/2. Then, yp(x)=(12)e3x, and the general solution is
y(x)=c1e−x+c2e2x+12e3x.
Find the general solution to y″−4y′+4y=7sint−cost.
- Hint
-
Use yp(t)=Asint+Bcost as a guess for the particular solution.
- Answer
-
y(t)=c1e2t+c2te2t+sint+cost
In the previous checkpoint, r(x) included both sine and cosine terms. However, even if r(x) included a sine term only or a cosine term only, both terms must be present in the guess. The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of r(x) and the associated guesses for yp(x) are summarized in Table 17.2.1.
r(x) | Initial guess for yp(x) |
---|---|
k (a constant) | A (a constant) |
ax+b | Ax+B (Note: The guess must include both terms even if b=0.) |
ax2+bx+c | Ax2+Bx+C (Note: The guess must include all three terms even if b or c are zero.) |
Higher-order polynomials | Polynomial of the same order as r(x) |
ae^{λx} | Ae^{λx} |
a \cos βx+b \sin βx | A \cos βx+B \sin βx (Note: The guess must include both terms even if either a=0 or b=0.) |
ae^{αx} \cos βx+be^{αx} \sin βx | Ae^{αx} \cos βx+Be^{αx} \sin βx |
(ax^2+bx+c)e^{λx} | (Ax^2+Bx+C)e^{λx} |
(a_2x^2+a_1x+a0) \cos βx \\ +(b_2x^2+b_1x+b_0) \sin βx | (A_2x^2+A_1x+A_0) \cos βx \\ +(B_2x^2+B_1x+B_0) \sin βx |
(a_2x^2+a_1x+a_0)e^{αx} \cos βx \\ +(b_2x^2+b_1x+b_0)e^{αx} \sin βx | (A_2x^2+A_1x+A_0)e^{αx} \cos βx \\ +(B_2x^2+B_1x+B_0)e^{αx} \sin βx |
Keep in mind that there is a key pitfall to this method. Consider the differential equation y″+5y′+6y=3e^{−2x}. Based on the form of r(x), we guess a particular solution of the form y_p(x)=Ae^{−2x}. But when we substitute this expression into the differential equation to find a value for A,we run into a problem. We have
y_p′(x)=−2Ae^{−2x} \nonumber
and
y_p''=4Ae^{−2x}, \nonumber
so we want
\begin{align*} y″+5y′+6y &=3e^{−2x} \\[4pt] 4Ae^{−2x}+5(−2Ae^{−2x})+6Ae^{−2x} &=3e^{−2x} \\[4pt] 4Ae^{−2x}−10Ae^{−2x}+6Ae^{−2x} &=3e^{−2x} \\[4pt] 0 &=3e^{−2x}, \end{align*}
which is not possible.
Looking closely, we see that, in this case, the general solution to the complementary equation is c_1e^{−2x}+c_2e^{−3x}. The exponential function in r(x) is actually a solution to the complementary equation, so, as we just saw, all the terms on the left side of the equation cancel out. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by x. Using the new guess, y_p(x)=Axe^{−2x}, we have
y_p′(x)=A(e^{−2x}−2xe^{−2x} \nonumber
and
y_p''(x)=−4Ae^{−2x}+4Axe^{−2x}. \nonumber
Substitution gives
\begin{align*}y″+5y′+6y &=3e^{−2x} \\[4pt] (−4Ae^{−2x}+4Axe^{−2x})+5(Ae^{−2x}−2Axe^{−2x})+6Axe^{−2x} &=3e^{−2x} \\[4pt]−4Ae^{−2x}+4Axe^{−2x}+5Ae^{−2x}−10Axe^{−2x}+6Axe^{−2x} &=3e^{−2x} \\[4pt] Ae^{−2x} &=3e^{−2x}.\end{align*}
So, A=3 and y_p(x)=3xe^{−2x}. This gives us the following general solution
y(x)=c_1e^{−2x}+c_2e^{−3x}+3xe^{−2x}. \nonumber
Note that if xe^{−2x} were also a solution to the complementary equation, we would have to multiply by x again, and we would try y_p(x)=Ax^2e^{−2x}.
- Solve the complementary equation and write down the general solution.
- Based on the form of r(x), make an initial guess for y_p(x).
- Check whether any term in the guess fory_p(x) is a solution to the complementary equation. If so, multiply the guess by x. Repeat this step until there are no terms in y_p(x) that solve the complementary equation.
- Substitute y_p(x) into the differential equation and equate like terms to find values for the unknown coefficients in y_p(x).
- Add the general solution to the complementary equation and the particular solution you just found to obtain the general solution to the nonhomogeneous equation.
Find the general solutions to the following differential equations.
- y''−9y=−6 \cos 3x
- x″+2x′+x=4e^{−t}
- y″−2y′+5y=10x^2−3x−3
- y''−3y′=−12t
Solution
- The complementary equation is y″−9y=0, which has the general solution c_1e^{3x}+c_2e^{−3x}(step 1). Based on the form of r(x)=−6 \cos 3x, our initial guess for the particular solution is y_p(x)=A \cos 3x+B \sin 3x (step 2). None of the terms in y_p(x) solve the complementary equation, so this is a valid guess (step 3).
Now we want to find values for A and B, so substitute y_p into the differential equation. We havey_p′(x)=−3A \sin 3x+3B \cos 3x \text{ and } y_p″(x)=−9A \cos 3x−9B \sin 3x, \nonumber
so we want to find values of A and B such that\begin{align*}y″−9y &=−6 \cos 3x \\[4pt] −9A \cos 3x−9B \sin 3x−9(A \cos 3x+B \sin 3x) &=−6 \cos 3x \\[4pt] −18A \cos 3x−18B \sin 3x &=−6 \cos 3x. \end{align*}
Therefore,\begin{align*}−18A &=−6 \\[4pt] −18B &=0. \end{align*}
This gives A=\frac{1}{3} and B=0, so y_p(x)=(\frac{1}{3}) \cos 3x (step 4).
Putting everything together, we have the general solutiony(x)=c_1e^{3x}+c_2e^{−3x}+\dfrac{1}{3} \cos 3x.\nonumber
- The complementary equation is x''+2x′+x=0, which has the general solution c_1e^{−t}+c_2te^{−t} (step 1). Based on the form r(t)=4e^{−t}, our initial guess for the particular solution is x_p(t)=Ae^{−t} (step 2). However, we see that this guess solves the complementary equation, so we must multiply by t, which gives a new guess: x_p(t)=Ate^{−t} (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by t again, which gives x_p(t)=At^2e^{−t} (step 3 again). Now, checking this guess, we see that x_p(t) does not solve the complementary equation, so this is a valid guess (step 3 yet again).
We now want to find a value for A, so we substitute x_p into the differential equation. We have\begin{align*}x_p(t) &=At^2e^{−t}, \text{ so} \\[4pt] x_p′(t) &=2Ate^{−t}−At^2e^{−t} \end{align*}
and x_p″(t)=2Ae^{−t}−2Ate^{−t}−(2Ate^{−t}−At^2e^{−t})=2Ae^{−t}−4Ate^{−t}+At^2e^{−t}. \nonumber
Substituting into the differential equation, we want to find a value of A so that\begin{align*} x″+2x′+x &=4e^{−t} \\[4pt] 2Ae^{−t}−4Ate^{−t}+At^2e^{−t}+2(2Ate^{−t}−At^2e^{−t})+At^2e^{−t} &=4e^{−t} \\[4pt] 2Ae^{−t}&=4e^{−t}. \end{align*} \nonumber
This gives A=2, so x_p(t)=2t^2e^{−t} (step 4). Putting everything together, we have the general solutionx(t)=c_1e^{−t}+c_2te^{−t}+2t^2e^{−t}.\nonumber
- The complementary equation is y″−2y′+5y=0, which has the general solution c_1e^x \cos 2x+c_2 e^x \sin 2x (step 1). Based on the form r(x)=10x^2−3x−3, our initial guess for the particular solution is y_p(x)=Ax^2+Bx+C (step 2). None of the terms in y_p(x) solve the complementary equation, so this is a valid guess (step 3). We now want to find values for A, B, and C, so we substitute y_p into the differential equation. We have y_p′(x)=2Ax+B and y_p″(x)=2A, so we want to find values of A, B, and C such that
\begin{align*}y″−2y′+5y &=10x^2−3x−3 \\[4pt] 2A−2(2Ax+B)+5(Ax^2+Bx+C) &=10x^2−3x−3 \\[4pt] 5Ax^2+(5B−4A)x+(5C−2B+2A) &=10x^2−3x−3. \end{align*}
Therefore,\begin{align*} 5A &=10 \\[4pt] 5B−4A &=−3 \\[4pt] 5C−2B+2A &=−3. \end{align*}
This gives A=2, B=1, and C=−1, so y_p(x)=2x^2+x−1 (step 4). Putting everything together, we have the general solutiony(x)=c_1e^x \cos 2x+c_2e^x \sin 2x+2x^2+x−1.\nonumber
- The complementary equation is y″−3y′=0, which has the general solution c_1e^{3t}+c_2 (step 1). Based on the form r(t)=−12t,r(t)=−12t, our initial guess for the particular solution is y_p(t)=At+B (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by t, which gives a new guess: y_p(t)=At^2+Bt (step 3). Checking this new guess, we see that none of the terms in y_p(t) solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for A and B, so we substitute y_p into the differential equation. We have y_p′(t)=2At+B and y_p″(t)=2A, so we want to find values of AA and BB such that
\begin{align*}y″−3y′ &=−12t \\[4pt] 2A−3(2At+B) &=−12t \\[4pt] −6At+(2A−3B) &=−12t. \end{align*}
Therefore,\begin{align*}−6A &=−12 \\[4pt] 2A−3B &=0. \end{align*}
This gives A=2 and B=4/3, so y_p(t)=2t^2+(4/3)t (step 4). Putting everything together, we have the general solutiony(t)=c_1e^{3t}+c_2+2t^2+\dfrac{4}{3}t.\nonumber
Find the general solution to the following differential equations.
- y″−5y′+4y=3e^x
- y″+y′−6y=52 \cos 2t
- Hint
-
Use the problem-solving strategy.
- Answer a
-
y(x)=c_1e^{4x}+c_2e^x−xe^x
- Answer b
-
y(t)=c_1e^{−3t}+c_2e^{2t}−5 \cos 2t+ \sin 2t
Variation of Parameters
Sometimes, r(x) 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.
To simplify our calculations a little, we are going to divide the differential equation through by a, so we have a leading coefficient of 1. Then the differential equation has the form
y″+py′+qy=r(x), \nonumber
where p and q are constants.
If the general solution to the complementary equation is given by c_1y_1(x)+c_2y_2(x), we are going to look for a particular solution of the form
y_p(x)=u(x)y_1(x)+v(x)y_2(x). \nonumber
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 x, rather than constants. We want to find functions u(x) and v(x) such that y_p(x) satisfies the differential equation. We have
\begin{align*}y_p &=uy_1+vy_2 \\[4pt] y_p′ &=u′y_1+uy_1′+v′y_2+vy_2′ \\[4pt] y_p″ &=(u′y_1+v′y_2)′+u′y_1′+uy_1″+v′y_2′+vy_2″. \end{align*}
Substituting into the differential equation, we obtain
\begin{align*}y_p″+py_p′+qy_p &=[(u′y_1+v′y_2)′+u′y_1′+uy_1″+v′y_2′+vy_2″] \\ &\;\;\;\;+p[u′y_1+uy_1′+v′y_2+vy_2′]+q[uy_1+vy_2] \\[4pt] &=u[y_1″+p_y1′+qy_1]+v[y_2″+py_2′+qy_2] \\ &\;\;\;\; +(u′y_1+v′y_2)′+p(u′y_1+v′y_2)+(u′y_1′+v′y_2′). \end{align*}
Note that y_1 and y_2 are solutions to the complementary equation, so the first two terms are zero. Thus, we have
(u′y_1+v′y_2)′+p(u′y_1+v′y_2)+(u′y_1′+v′y_2′)=r(x). \nonumber
If we simplify this equation by imposing the additional condition u′y_1+v′y_2=0, the first two terms are zero, and this reduces to u′y_1′+v′y_2′=r(x). So, with this additional condition, we have a system of two equations in two unknowns:
\begin{align*} u′y_1+v′y_2 &= 0 \\[4pt] u′y_1′+v′y_2′ &=r(x). \end{align*}
Solving this system gives us u′ and v′, which we can integrate to find u and v.
Then, y_p(x)=u(x)y_1(x)+v(x)y_2(x) is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let’s take this opportunity to review Cramer’s rule, which allows us to solve the system of equations using determinants.
The system of equations
\begin{align*} a_1z_1+b_1z_2 &=r_1 \\[4pt] a_2z_1+b_2z_2 &=r_2 \end{align*}
has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by
z_1=\dfrac{\begin{array}{|ll|}r_1 b_1 \\ r_2 b_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}} \; \; \; \; \; \text{and} \; \; \; \; \; z_2= \dfrac{\begin{array}{|ll|}a_1 r_1 \\ a_2 r_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}}. \label{cramer}
Use Cramer’s rule to solve the following system of equations.
\begin{align*}x^2z_1+2xz_2 &=0 \\[4pt] z_1−3x^2z_2 &=2x \end{align*}
Solution
We have
\begin{align*} a_1(x) &=x^2 \\[4pt] a_2(x) &=1 \\[4pt] b_1(x) &=2x \\[4pt] b_2(x) &=−3x^2 \\[4pt] r_1(x) &=0 \\[4pt] r_2(x) &=2x. \end{align*}
Then,\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}=\begin{array}{|ll|}x^2 2x \\ 1 −3x^2 \end{array}=−3x^4−2x \nonumber
and
\begin{array}{|ll|}r_1 b_1 \\ r_2 b_2 \end{array}=\begin{array}{|ll|}0 2x \\ 2x -3x^2 \end{array}=0−4x^2=−4x^2. \nonumber
Thus,
z1=\dfrac{\begin{array}{|ll|}r_1 b_1 \\ r_2 b_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}}=\dfrac{−4x^2}{−3x^4−2x}=\dfrac{4x}{3x^3+2}. \nonumber
In addition,
\begin{array}{|ll|}a_1 r_1 \\ a_2 r_2 \end{array}=\begin{array}{|ll|} x^2 0 \\ 1 2x \end{array}=2x^3−0=2x^3. \nonumber
Thus,
z2=\dfrac{\begin{array}{|ll|}a_1 r_1 \\ a_2 r_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}}=\dfrac{2x^3}{−3x^4−2x}=\dfrac{−2x^2}{3x^3+2}.\nonumber
Use Cramer’s rule to solve the following system of equations.
\begin{align*} 2xz_1−3z_2 &=0 \\[4pt] x^2z_1+4xz_2 &=x+1 \end{align*}
- Hint
-
Use the process from the previous example.
- Answer
-
z_1=\frac{3x+3}{11x^2}, z_2=\frac{2x+2}{11x}
- Solve the complementary equation and write down the general solution c_1y_1(x)+c_2y_2(x). \nonumber
- Use Cramer’s rule or another suitable technique to find functions u′(x) and v′(x) satisfying \begin{align*} u′y_1+v′y_2 &=0 \\[4pt] u′y_1′+v′y_2′ &=r(x). \end{align*}
- Integrate u′ and v′ to find u(x) and v(x). Then, y_p(x)=u(x)y_1(x)+v(x)y_2(x) is a particular solution to the equation.
- Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.
Find the general solution to the following differential equations.
- y″−2y′+y=\dfrac{e^t}{t^2}
- y″+y=3 \sin ^2 x
Solution
- The complementary equation is y″−2y′+y=0 with associated general solution c_1e^t+c_2te^t. Therefore, y_1(t)=e^t and y_2(t)=te^t. Calculating the derivatives, we get y_1′(t)=e^t and y_2′(t)=e^t+te^t (step 1). Then, we want to find functions u′(t) and v′(t) so that
\begin{align*} u′e^t+v′te^t &=0 \\[4pt] u′e^t+v′(e^t+te^t) &= \dfrac{e^t}{t^2}. \end{align*}
Applying Cramer’s rule (Equation \ref{cramer}), we have
u′=\dfrac{\begin{array}{|lc|}0 te^t \\ \frac{e^t}{t^2} e^t+te^t \end{array}}{ \begin{array}{|lc|}e^t te^t \\ e^t e^t+te^t \end{array}} =\dfrac{0−te^t(\frac{e^t}{t^2})}{e^t(e^t+te^t)−e^tte^t}=\dfrac{−\frac{e^{2t}}{t}}{e^{2t}}=−\dfrac{1}{t} \nonumber
and
v′= \dfrac{\begin{array}{|ll|}e^t 0 \\ e^t \frac{e^t}{t^2} \end{array} }{\begin{array}{|lc|}e^t te^t \\ e^t e^t+te^t \end{array} } =\dfrac{e^t(\frac{e^t}{t^2})}{e^{2t}}=\dfrac{1}{t^2}\quad(\text{step 2}). \nonumber
Integrating, we get
\begin{align*} u &=−\int \dfrac{1}{t}dt=− \ln|t| \\[4pt] v &=\int \dfrac{1}{t^2}dt=−\dfrac{1}{t} \tag{step 3} \end{align*}
Then we have
\begin{align*}y_p &=−e^t \ln|t|−\frac{1}{t}te^t \\[4pt] &=−e^t \ln |t|−e^t \tag{step 4}.\end{align*}
The e^t term is a solution to the complementary equation, so we don’t need to carry that term into our general solution explicitly. The general solution is
y(t)=c_1e^t+c_2te^t−e^t \ln |t| \tag{step 5}
- The complementary equation is y″+y=0 with associated general solution c_1 \cos x+c_2 \sin x. So, y_1(x)= \cos x and y_2(x)= \sin x (step 1). Then, we want to find functions u′(x) and v′(x) such that
\begin{align*} u′ \cos x+v′ \sin x &=0 \\[4pt] −u′ \sin x+v′ \cos x &=3 \sin _2 x \end{align*}. \nonumber
Applying Cramer’s rule, we have
u′= \dfrac{\begin{array}{|cc|}0 \sin x \\ 3 \sin ^2 x \cos x \end{array}}{ \begin{array}{|cc|} \cos x \sin x \\ − \sin x \cos x \end{array}}=\dfrac{0−3 \sin^3 x}{ \cos ^2 x+ \sin ^2 x}=−3 \sin^3 x \nonumber
and
v′=\dfrac{\begin{array}{|cc|} \cos x 0 \\ - \sin x 3 \sin^2 x \end{array}}{ \begin{array}{|cc|} \cos x \sin x \\ − \sin x \cos x \end{array}}=\dfrac{ 3 \sin^2x \cos x}{ 1}=3 \sin^2 x \cos x( \text{step 2}). \nonumber
Integrating first to find u, we get
u=\int −3 \sin^3 x dx=−3 \bigg[ −\dfrac{1}{3} \sin ^2 x \cos x+\dfrac{2}{3} \int \sin x dx \bigg]= \sin^2 x \cos x+2 \cos x. \nonumber
Now, we integrate to find v. Using substitution (with w= \sin x), we get
v= \int 3 \sin ^2 x \cos x dx=\int 3w^2dw=w^3=sin^3x.\nonumber
Then,
\begin{align*}y_p &=(\sin^2 x \cos x+2 \cos x) \cos x+(\sin^3 x)\sin x \\[4pt] &=\sin_2 x \cos _2 x+2 \cos _2 x+ \sin _4x \\[4pt] &=2 \cos_2 x+ \sin_2 x(\cos^2 x+\sin ^2 x) & & (\text{step 4}). \\[4pt] &=2 \cos _2 x+\sin_2x \\[4pt] &= \cos _2 x+1 \end{align*}
The general solution is
y(x)=c_1 \cos x+c_2 \sin x+1+ \cos^2 x(\text{step 5}).\nonumber
Find the general solution to the following differential equations.
- y″+y= \sec x
- x″−2x′+x=\dfrac{e^t}{t}
- Hint
-
Follow the problem-solving strategy.
- Answer a
-
y(x)=c_1 \cos x+c_2 \sin x+ \cos x \ln| \cos x|+x \sin x
- Answer b
-
x(t)=c_1e^t+c_2te^t+te^t \ln|t|
Key Concepts
- To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation.
- Let y_p(x) be any particular solution to the nonhomogeneous linear differential equation a_2(x)y''+a_1(x)y′+a_0(x)y=r(x), \nonumber and let c_1y_1(x)+c_2y_2(x) denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by y(x)=c_1y_1(x)+c_2y_2(x)+y_p(x). \nonumber
- When r(x) is a combination of polynomials, exponential functions, sines, and cosines, use the method of undetermined coefficients to find the particular solution. To use this method, assume a solution in the same form as r(x), multiplying by x as necessary until the assumed solution is linearly independent of the general solution to the complementary equation. Then, substitute the assumed solution into the differential equation to find values for the coefficients.
- When r(x) is not a combination of polynomials, exponential functions, or sines and cosines, use the method of variation of parameters to find the particular solution. This method involves using Cramer’s rule or another suitable technique to find functions and v′(x) satisfying \begin{align*}u′y_1+v′y_2 &=0 \\[4pt] u′y_1′+v′y_2′ &=r(x). \end{align*} Then, y_p(x)=u(x)y_1(x)+v(x)y_2(x) is a particular solution to the differential equation.
Key Equations
- Complementary equation
a_2(x)y″+a_1(x)y′+a_0(x)y=0 - General solution to a nonhomogeneous linear differential equation
y(x)=c_1y_1(x)+c_2y_2(x)+y_p(x)
Glossary
- complementary equation
- for the nonhomogeneous linear differential equation a+2(x)y″+a_1(x)y′+a_0(x)y=r(x), the associated homogeneous equation, called the complementary equation, is a_2(x)y''+a_1(x)y′+a_0(x)y=0
- method of undetermined coefficients
- a method that involves making a guess about the form of the particular solution, then solving for the coefficients in the guess
- method of variation of parameters
- a method that involves looking for particular solutions in the form y_p(x)=u(x)y_1(x)+v(x)y_2(x), where y_1 and y_2 are linearly independent solutions to the complementary equations, and then solving a system of equations to find u(x) and v(x)
- particular solution
- a solution y_p(x) of a differential equation that contains no arbitrary constants