Question

5. Apply the method of variations of parameters to solve the following differential equations:

a) x^2 y'' + x y' - y = x^2 e^x

b) y'' + a^2 y = cosec ax

c) Solve the equation d^2 y / d x^2 - cot x dy/dx - sin^2 xy = cos x - cos^3 x by changing the independent
variable.

Answer 1

Answer on Question #50899-Math-Differential Calculus-Equations

Apply the method of variations of parameters to solve the following differential equations:

a) $x^{2}y^{\prime \prime} + xy^{\prime} - y = x^{2}e^{x}$

b) $y^{\prime \prime} + a^{2}y = \text{cosec} ax$

c) Solve the equation $\frac{d^2y}{dx^2} - \cot x\frac{dy}{dx} - \sin^2 xy = \cos x - \cos^3 x$ by changing the independent variable.

Solution

a) $x^{2}y^{\prime \prime} + xy^{\prime} - y = 0$

$$\begin{array}{l} y = x^{n} \rightarrow n(n - 1) + n - 1 = 0 \rightarrow n = \pm 1. \\ y = c_{1}x + c_{2}\frac{1}{x}. \end{array}$$

The Wronskian is given by

$$W = \begin{bmatrix} x & \frac{1}{x} \\ 1 & -\frac{1}{x^{2}} \end{bmatrix} = -\left(\frac{1}{x} + \frac{1}{x}\right) = -\frac{2}{x} \neq 0.$$

We seek the solution in the form $y = y_{1}v_{1} + y_{2}v_{2}$

$$\begin{array}{l} v_{1} = \int \left(-\frac{y_{2}R}{W}\right) dx = -\int \frac{\frac{1}{x} \cdot x^{2} e^{x}}{-\frac{2}{x}} dx = \frac{e^{x}}{2} (x^{2} - 2x + 2) + c_{1}. \\ v_{2} = \int \left(\frac{y_{1}R}{W}\right) dx = \int \frac{x \cdot x^{2} e^{x}}{-\frac{2}{x}} dx = -\frac{e^{x}}{2} (x^{4} - 4x^{3} + 12x^{2} - 24x + 24) + c_{2}. \end{array}$$

So,

$$\begin{array}{l} y = x \left[ \frac{e^{x}}{2} (x^{2} - 2x + 2) + c_{1} \right] + \frac{1}{x} \left[ -\frac{e^{x}}{2} (x^{4} - 4x^{3} + 12x^{2} - 24x + 24) + c_{2} \right]. \\ y = c_{1}x + c_{2}\frac{1}{x} + e^{x} \left[ x^{2} - 5x + 12 - \frac{12}{x} \right] \end{array}$$

b) $y^{\prime \prime} + a^{2}y = 0$

$$y = c_{2}\cos ax + c_{1}\sin ax.$$

The Wronskian is given by

$$W = \begin{bmatrix} \sin ax & \cos ax \\ acos ax & -asin ax \end{bmatrix} = -a(\sin^{2} ax + \cos^{2} ax) = -a \neq 0.$$

We seek the solution in the form $y = y_{1}v_{1} + y_{2}v_{2}$

$$v_{1} = \int \left(-\frac{y_{2}R}{W}\right) dx = -\int \frac{\cos ax \cdot \text{cosec} ax}{-a} dx = \frac{1}{a^{2}} \ln |\sin ax| + c_{1}.$$ $$v _ {2} = \int \left(\frac {y _ {1} R}{W}\right) d x = \int \frac {\sin a x \cdot \csc a x}{- a} d x = \frac {x}{a} + c _ {2}.$$

So,

$$y = \cos a x \left[ \frac {x}{a} + c _ {2} \right] + \sin a x \left[ \frac {1}{a ^ {2}} \ln | \sin a x | + c _ {1} \right].$$

c) Comparing the given equation with $y'' + Py' + Qy = R$, we have $P = -\cot x$, $Q = -\sin^2 x$ and $R = \cos x - \cos^3 x = \cos x \sin^2 x$. (1)

Choose $z$ such that $\left(\frac{dz}{dx}\right)^2 = \sin^2 x$ and $\frac{dz}{dx} = \sin x$. (2)

Integrating, $z = \int \sin x \, dx = -\cos x$. (3)

Now changing the independent variable from $x$ to $z$ by using relation (3), the given equation is transformed into

$$\frac {d ^ {2} y}{d z ^ {2}} + P _ {1} \frac {d y}{d z} + Q _ {1} y = R _ {1},$$

where $P_{1} = \frac{\frac{d^{2}z}{dx^{2}} + P\frac{dz}{dx}}{\left(\frac{dz}{dx}\right)^{2}} = \frac{\cos x + (-cot x)(\sin x)}{\sin^{2}x} = 0$, by (1) and (2)

$$Q _ {1} = \frac {Q}{\left(\frac {d z}{d z}\right) ^ {2}} = \frac {- \sin^ {2} x}{\sin^ {2} x} = - 1, R _ {1} = \frac {R}{\left(\frac {d z}{d z}\right) ^ {2}} = \frac {\cos x \sin^ {2} x}{\sin^ {2} x} = \cos x = - z.$$

From (5), $\frac{d^2y}{dz^2} - y = -z$, or $(D_1^2 - 1)y = -z$, where $D_1 = \frac{d}{dz}$.

Its auxiliary equation is $m^2 - 1 = 0$ so that $m = \pm 1$. General solution of (5) is

$$c _ {1} e ^ {z} + c _ {2} e ^ {- z} = c _ {1} e ^ {- \cos x} + c _ {2} e ^ {\cos x}, \text{ by (3)}$$

Partial solution is

$$\frac {1}{D _ {1} ^ {2} - 1} (- z) = \frac {1}{1 - D _ {1} ^ {2}} z = (1 - D _ {1} ^ {2}) ^ {- 1} z = (1 + D _ {1} ^ {2} + \dots) z = z = - \cos x.$$

Hence the required solution is

$$y = c _ {1} e ^ {- \cos x} + c _ {2} e ^ {\cos x} - \cos x.$$

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