17.2: Nonhom*ogeneous Linear Equations (2024)

Solution

The complementary equation is \(y″+y=0,\) which has the general solution \(c_1 \cos x+c_2 \sin x.\) So, the general solution to the nonhom*ogeneous equation is

\[y(x)=c_1 \cos x+c_2 \sin x+x. \nonumber \]

To verify that this is a solution, substitute it into the differential equation. We have

\[y′(x)=−c_1 \sin x+c_2 \cos x+1 \nonumber \]

and

\[y″(x)=−c_1 \cos x−c_2 \sin x. \nonumber \]

Then

\[\begin{align*} y″(x)+y(x) &=−c_1 \cos x−c_2 \sin x+c_1 \cos x+c_2 \sin x+x \\[4pt] &=x.\end{align*} \nonumber \]

So, \(y(x)\) is a solution to \(y″+y=x\).

Solution

The complementary equation is \(y″+4y′+3y=0\), with general solution \(c_1e^{−x}+c_2e^{−3x}\). Since \(r(x)=3x\), the particular solution might have the form \(y_p(x)=Ax+B\). If this is the case, then we have \(y_p′(x)=A\) and \(y_p″(x)=0\). For \(y_p\) to be a solution to the differential equation, we must find values for \(A\) and \(B\) such that

\[\begin{align*} y″+4y′+3y &=3x \\[4pt] 0+4(A)+3(Ax+B) &=3x \\[4pt] 3Ax+(4A+3B) &=3x. \nonumber \end{align*} \nonumber \]

Setting coefficients of like terms equal, we have

\[\begin{align*} 3A &=3 \\ 4A+3B &=0. \end{align*} \nonumber \]

Then, \(A=1\) and \(B=−\frac{4}{3}\), so \(y_p(x)=x−\frac{4}{3}\) and the general solution is

\[y(x)=c_1e^{−x}+c_2e^{−3x}+x−\frac{4}{3}. \nonumber \]

Solution

The complementary equation is \(y″−y′−2y=0\), with the general solution \(c_1e^{−x}+c_2e^{2x}\). Since \(r(x)=2e^{3x}\), the particular solution might have the form \(y_p(x)=Ae^{3x}.\) Then, we have \(yp′(x)=3Ae^{3x}\) and \(y_p″(x)=9Ae^{3x}\). For \(y_p\) to be a solution to the differential equation, we must find a value for \(A\) such that

\[\begin{align*} y″−y′−2y &=2e^{3x} \\[4pt] 9Ae^{3x}−3Ae^{3x}−2Ae^{3x} &=2e^{3x} \\[4pt] 4Ae^{3x} &=2e^{3x}. \end{align*} \nonumber \]

So, \(4A=2\) and \(A=1/2\). Then, \(y_p(x)=(\frac{1}{2})e^{3x}\), and the general solution is

\[y(x)=c_1e^{−x}+c_2e^{2x}+\dfrac{1}{2}e^{3x}. \nonumber \]

Solution

1. 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 have

   \[y_p′(x)=−3A \sin 3x+3B \cos 3x \text{ and } y_p″(x)=−9A \cos 3x−9B \sin 3x, \nonumber \]

   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 solution

   \[y(x)=c_1e^{3x}+c_2e^{−3x}+\dfrac{1}{3} \cos 3x.\nonumber \]

2. 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 solution

   \[x(t)=c_1e^{−t}+c_2te^{−t}+2t^2e^{−t}.\nonumber \]

3. 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 solution

   \[y(x)=c_1e^x \cos 2x+c_2e^x \sin 2x+2x^2+x−1.\nonumber \]

4. 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 solution

   \[y(t)=c_1e^{3t}+c_2+2t^2+\dfrac{4}{3}t.\nonumber \]

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,

\[z2=\dfrac{\begin{array}{|ll|}a_1 r_1 \\ a_2 r_2 \end{array}}{\begin{array}{|ll|}a_1 b_1 \\ a_2 b_2 \end{array}}=\dfrac{2x^3}{−3x^4−2x}=\dfrac{−2x^2}{3x^3+2}.\nonumber \]

Solution

1. 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

   \[y(t)=c_1e^t+c_2te^t−e^t \ln |t| \tag{step 5} \]

2. 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

   \[y(x)=c_1 \cos x+c_2 \sin x+1+ \cos^2 x(\text{step 5}).\nonumber \]