Variation of Parameter

Exercise 2.5

1. Solve the following differential equation using variation of parameter.

(a) $\ y^{\prime\prime} + 2y^{\prime} + y = \frac{e^{-x}}{x}$
(b) $\ y^{\prime\prime} + 4y = \tan{2x}$
(c) $\ y^{\prime\prime} - 4y^{\prime} + 4y = \frac{e^{2x}}{x}$
(d) $\ y^{\prime\prime} - 3y^{\prime} + 2y = \frac{1}{1+e^{-x}}$
(e) $\ y^{\prime\prime\prime} + y^{\prime} = \tan{x}$

2.
Show that using variation of parameter the general solution of $y^{\prime\prime} + y = f(x)$ is given by

$$y = c_{1}\cos{x} + c_{2}\sin{x} + \int_{a}^{x}f(t)\sin{(x-t)}dt$$

Answer
1. (a) Given auxiliary equation $y^{\prime\prime} + 2y^{\prime} + y = 0$. The characteristic equation is $m^2 + 2m + 1 = (m + 1)^2 = 0$ and characteristic roots are $m = -1, -1$. Thus the complementary function is

$$y_{c} = (c_{1} + c_{2}x)e^{-x}$$

We next find the particular solution using the variation of parameter. Set

$$y_{p} = u_{1}e^{-x} + u_{2}xe^{-x}$$

Then we have

$$\left\{\begin{array}{ll} u_{1}^{\prime}e^{-x} + u_{2}^{\prime}xe^... ...x}) + u_{2}^{\prime}(e^{-x} - xe^{-x}) & = \frac{e^{-x}}{x} \end{array}\right.$$

Solve this using Cramer's rule. Then we have

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0 & xe^{-x}\\ \frac{e^{-x}}{x}... ...-x} & e^{-x} - xe^{-x} \end{array}\right\vert } = \frac{-e^{-2x}}{e^{-2x}} = -1$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} e^{-x} & 0\\ -e^{-x} & \frac{e... ...{x} \end{array}\right\vert }{e^{-2x}} = \frac{e^{-2x}/x}{e^{-2x}} = \frac{1}{x}$$

Thus,

$$u_{1} = -x, \ u_{2} = \log{\vert x\vert} \$$   (ignore constant)$$$$

Thus,

$$y_{p} = -xe^{-x} - x\log{\vert x\vert}e^{-x}.$$

Note that $xe^{-x}$ is already used in $y_c$. So, we omit from $y_p$. Then the general solution is

$$y = y_{c} + y_{p} = (c_{1} + c_{2}x)e^{-x} - x\log{\vert x\vert}e^{-x}$$

(b) Given auxiliary equation $y^{\prime\prime} + 4y = 0$. Then the characteristic equation is $m^2 + 4 = 0$ and the characteristic roots are $m = \pm 2i$. Then the complementary function is given by

$$y_{c} = c_{1}\cos{2x} + c_{2}\sin{2x}$$

Next we find the particular solution using the variation of parameter. Let

$$y_{p} = u_{1}\cos{2x} + u_{2}\sin{2x}$$

Then we have

$$\left\{\begin{array}{ll} u_{1}^{\prime}\cos{2x} + u_{2}^{\prime}\... ...prime}(-2\sin{2x}) + u_{2}^{\prime}(2\cos{2x}) & = \tan{2x} \end{array}\right.$$

Now solve these equations using Cramer's rule.

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0 & \sin{2x}\\ \tan{2x} & 2\co... ...tan{2x}\sin{2x}}{2(\cos^{2}{2x} + \sin^{2}{2x})} = - \frac{\tan{2x}\sin{2x}}{2}$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} \cos{2x} & 0\\ -2\sin{2x} & \tan{2x} \end{array}\right\vert }{2} = \frac{\sin{2x}}{2}$$

Now find $u_{1},u_{2}$.

$$u_{1} = -\frac{1}{2}\int \frac{\sin^{2}{2x}}{\cos{2x}} = \frac{1}{2}\int \frac{\cos^{2}{2x} - 1}{\cos{2x}} dx$$

$$= \frac{1}{2}\int (\cos{2x} - \sec{2x})dx$$

$$= -\frac{1}{4}(\sin{2x} + \log{\vert\sec{2x} + \tan{2x}\vert})$$

$$u_{2} = \int \frac{\sin{2x}}{2}dx = - \frac{\cos{2x}}{4}$$

Thus,

$$y_{p} = -\frac{1}{4}\cos{2x}( \sin{2x} + \log{\vert\sec{2x} + \tan{2x}\vert}) - \frac{1}{4}\cos{2x} \sin{2x} .$$

Therefore, the general solution is

$$y = y_{c} + y_{p} = c_{1}\cos{2x} + c_{2}\sin{2x}$$

$$- \frac{1}{4}\cos{2x} \log{\vert\sec{2x} + \tan{2x}\vert} - \frac{1}{2}\sin{2x}\cos{2x}$$

(c) Given auxiliary equation $y^{\prime\prime} - 4y^{\prime} + 4y = 0$. The characteristic equation is $m^2 - 4m + 4 = (m -2)^2 = 0$ and the characteristic roots are $m = 2, 2$. Thus, the complementary solution is give by

$$y_{c} = (c_{1} + c_{2}x)e^{2x}$$

Next we find the particular solution. Using the variation of parameter, we set

$$y_{p} = u_{1}e^{2x} + u_{2}xe^{2x}$$

Then we have

$$\left\{\begin{array}{ll} u_{1}^{\prime}e^{2x} + u_{2}^{\prime}xe^... ...}) + u_{2}^{\prime}(e^{2x} + 2xe^{2x}) & = \frac{e^{2x}}{x} \end{array}\right.$$

Solve this equation using Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0 & xe^{2x}\\ \frac{e^{2x}}{x}... ...x} & e^{2x} + 2xe^{2x} \end{array}\right\vert } = - \frac{e^{4x}}{e^{4x}} = - 1$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} e^{2x} & 0\\ 2e^{2x} & \frac{e... ...{2x} + 2xe^{2x} \end{array}\right\vert} = \frac{e^{4x}/x}{e^{4x}} = \frac{1}{x}$$

Thus, we have $u_{1} = -x, u_{2} = \log{\vert x\vert}$ and

$$y_{p} = - xe^{2x} + xe^{2x}\log{\vert x\vert}.$$

Note that $- xe^{2x}$ is already used in the complementary function. So, the general solution is

$$y = y_{c} + y_{p} = (c_{1} + c_{2}x)e^{2x} + xe^{2x}\log{\vert x\vert}$$

(d) Given auxiliary equation $y^{\prime\prime} - 3y^{\prime} + 2y = 0$, we ahve the characteristic equation which is $m^2 - 3m + 2 = (m -1)(m-2) = 0$. Then the characteristic roots are $m = 1, \ 2$. Thus the complementary function is give by

$$y_{c} = c_{1}e^{x} + c_{2}e^{2x}$$

We next find the particular solution. By the variation of parameter, we set

$$y_{p} = u_{1}e^{x} + u_{2}e^{2x}$$

Then we have

$$\left\{\begin{array}{ll} u_{1}^{\prime}e^{x} + u_{2}^{\prime}e^{2... ...}(e^{x}) + u_{2}^{\prime}(2e^{2x}) & = \frac{1}{1 + e^{-x}} \end{array}\right.$$

Solve this equatios by Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0 & e^{2x}\\ \frac{1}{1 + e^{-... ...ght\vert } = - \frac{e^{2x}/(1 + e^{-x})}{e^{3x}} = - \frac{e^{-x}}{1 + e^{-x}}$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} e^{x} & 0\\ e^{x} & \frac{1}{1... ...y}\right\vert} = \frac{e^{x}/(1 + e^{-x})}{e^{3x}} = \frac{e^{-2x}}{1 + e^{-x}}$$

We find $u_{1},u_{2}$. Then

$$u_{1} = -\int \frac{e^{-x}}{1 + e^{-x}} dx = \log{(1 + e^{-x})}$$

$$u_{2} = \int \frac{e^{-2x}}{1 + e^{-x}} dx = \int (e^{-x} - \frac{e^{-x}}{1 + e^{-x}})dx$$

$$= - e^{-x} + \log{(1 + e^{-x})}$$

Thus

$$y_{p} = e^{x} \log{(1 + e^{-x})} + e^{2x}(- e^{-x} + \log{(1 + e^{-x}})).$$

Therefore, the general solution is

$$y = y_{c} + y_{p} = c_{1}e^{x} + c_{2}e^{2x} + (e^{x} + e^{2x})\log{(1 + e^{-x})}$$

(e) The auxiliary equation is $y^{\prime\prime\prime} + y^{\prime} = 0$. Then the characteristic equation is $m^3 + m = m(m^2 + 1) = 0$ and characteristic roots are $m = 0, \pm i$. Thus, the complementary function is given by

$$y_{c} = c_{1} + c_{2}\cos{x} + c_{3}\sin{x}$$

We next find the particular solution. Using the variation of parameter, we set

$$y_{p} = u_{1} + u_{2}\cos{x} + u_{3}\sin{x}$$

Then we have

$$\left\{\begin{array}{ll} u_{1}^{\prime} + u_{2}^{\prime}\cos{x} +... ...}^{\prime}(-\cos{x}) + u_{3}^{\prime}(-\sin{x}) & = \tan{x} \end{array}\right.$$

Solve these equations using Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{ccc} 0 & \cos{x} & \sin{x}\\ 0 & -... ...in{x} & \cos{x}\\ 0 & - \cos{x} & - \sin{x} \end{array}\right\vert } = \tan{x}$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{ccc} 1 & 0 & \sin{x}\\ 0 & 0 & \co... ...& - \cos{x} & - \sin{x} \end{array}\right\vert } = - \tan{x}\cos{x} = - \sin{x}$$

$$u_{3}^{\prime} = \frac{\left\vert\begin{array}{ccc} 1 & \cos{x} & 0\\ 0 & - \sin{... ...cos{x}\\ 0 & - \cos{x} & - \sin{x} \end{array}\right\vert } = - \sin{x}\tan{x}$$

From this, we find $u_{1} , u_{2}, u_{3}$. Then

$$u_{1} = \int \tan{x} dx = \int \frac{\sin{x}}{\cos{x}}dx = - \log{\vert\cos{x}\vert} = \log{\vert\sec{x}\vert}$$

$$u_{2} = - \int \sin{x}dx = \cos{x}$$

$$u_{3} = - \int \sin{x}\tan{x}dx = - \int \frac{\sin^{2}{x}}{\cos{x}} dx = \int \frac{\cos^{2}{x} - 1}{\cos{x}}dx$$

$$= \int (\cos{x} - \sec{x})dx = \sin{x} - \log{\vert\sec{x} + \tan{x}\vert}$$

Thus,

$$y_{p} = \log{\vert\sec{x}\vert} + \cos^{2}{x} + \sin{x}(\sin{x} - \log{\vert\sec{x} + \tan{x}\vert})$$

$$= \log{\vert\sec{x}\vert} - \sin{x} \log{\vert\sec{x} + \tan{x}\vert} + 1$$

Therefore, the general solution is given by

$$y = y_{c} + y_{p} = c_{1} + c_{2}\cos{x} + c_{3}\sin{x} + \log{\vert\sec{x}\vert} - \sin{x} \log{\vert\sec{x} + \tan{x}\vert}$$

2. The auxiliary equation is $y^{\prime\prime} + y = 0$. Then the characterisitc equation is $m^2 + 1 = 0$ and characteristic roots are $m = \pm i$. Thus the complementary function is given by

$$y_{c} = c_{1}\cos{x} + c_{2}\sin{x}$$

We next find the particular solution. Using the variation of parameter, we set

$$y_{p} = u_{1}\cos{x} + u_{2}\sin{x}$$

Then we have

$$\left\{\begin{array}{ll} u_{1}^{\prime}\cos{x} + u_{2}^{\prime}\s... ...u_{1}^{\prime}(-\sin{x}) + u_{2}^{\prime}(\cos{x}) & = f(x) \end{array}\right.$$

Solve this using Cramer's rule. Then

$$u_{1}^{\prime} = \frac{\left\vert\begin{array}{cc} 0 & \sin{x}\\ f(x) & \cos{x} \... ...os{x} & \sin{x}\\ - \sin{x} & \cos{x} \end{array}\right\vert } = - f(x)\sin{x}$$

$$u_{2}^{\prime} = \frac{\left\vert\begin{array}{cc} \cos{x} & 0\\ - \sin{x} & f(x)... ...\cos{x} & \sin{x}\\ - \sin{x} & \cos{x} \end{array}\right\vert } = f(x)\cos{x}$$

Then we find $u_{1},u_{2}$.

$$u_{1} = - \int f(x)\sin{x} dx = - \int_{a}^{x}f(t)\sin{t}dt$$

$$u_{2} = \int f(x)\cos{x}dx = \int_{a}^{x}f(t)\cos(t)dt$$

Thus,

$$y_{p} = - \int_{a}^{x}f(t)\sin{t}dt \cos{x} + \int_{a}^{x}f(t)\cos(t)dt \sin{x}$$

$$= - \int_{a}^{x}f(t)\sin{t} \cos{x} dt + \int_{a}^{x}f(t)\cos(t) \sin{x} dt$$

$$= \int_{a}^{x} f(t)(\sin{x}\cos{t} - \cos{x}\sin{t})dt = \int_{a}^{x} f(t)(\sin{x}\cos{t} - \cos{x}\sin{t})dt$$

$$= \int_{a}^{x} f(t)\sin{(x - t)}dt$$

Therefore, the general solution is given by

$$y = y_{c} + y_{p} = c_{1}\cos{x} + c_{2}\sin{x} + \int_{a}^{x} f(t)\sin{(x - t)}dt$$