Question

Using method of variation of parameters, solve the equation.
d^2y/dx^2+y=Cosec x

Answer 1

Answer on Question #84315 – Math – Differential Equations

Question

Using method of variation of parameters, solve the equation.

$$d^2 y/dx^2 + y = \text{Cosec} \times$$

Solution

We have the equation: $y'' + y = \text{cosec}(x)$, or $y'' + y = \frac{1}{\sin(x)}$.

Solve the homogeneous equation: $y'' + y = 0$. We solve its characteristic equation: $k^2 + 1 = 0$, $k_1 = 1$, $k_2 = -1$. The general solution to the equation will be: $y^* = C_1 \cos(x) + C_2 \sin(x)$.

We vary the parameters. $y^* = C_1(x) \cos(x) + C_2(x) \sin(x)$.

We make the system of equations:

$$\begin{array}{l} C_1'(x) \cos(x) + C_2'(x) \sin(x) = 0, \\ -C_1'(x) \sin(x) + C_2'(x) \cos(x) = \frac{1}{\sin(x)}. \end{array}$$

From the first equation of the system we get the following: $C_1'(x) \cos(x) = -C_2'(x) \sin(x)$, $C_1'(x) = -C_2'(x) \tan(x)$. Substitute $C_1'(x)$ in the second equation of the system. We get:

$$\begin{array}{l} C_2'(x) \tan(x) \sin(x) + C_2'(x) \cos(x) = \frac{1}{\sin(x)}. \quad C_2'(x) = \frac{\cos(x)}{\sin(x) \times ((\cos(x))^2 - (\sin(x))^2)}, \quad \text{or} \\ d(C_2(x)) = \frac{\cos(x) \, dx}{\sin(x) \times ((\cos(x))^2 - (\sin(x))^2)}. \end{array}$$

We integrate the extreme expression. We get: $\int d(C_2'(x)) = \int \frac{\cos(x) \, dx}{\sin(x) \times ((\cos(x))^2 - (\sin(x))^2)}$, or

$$\begin{array}{l} C_2(x) = \int \frac{\cos(x) \, dx}{\sin(x) \times ((\cos(x))^2 - (\sin(x))^2)} = |\sin(x)| = t, \quad \cos(x) \, dx = dt, \quad (\cos(x))^2 - (\sin(x))^2 = 1 - 2t^2 = \\ = \int \frac{dt}{t(1 - 2t^2)} \, dt + \frac{Bt + C}{1 - 2t^2} \equiv \frac{1}{t(1 - 2t^2)}, \quad B - 2A = 0, \quad C = 0, \quad A = 1, \quad B = 2 = \int \frac{dt}{t} + \int \frac{2t \, dt}{1 - 2t^2} \, dt - \frac{1}{2} \ln |1 - 2t^2| + C_2^* = \\ = \ln |\sin(x)| - \frac{1}{2} \ln |1 - 2(\sin(x))^2| + C_2. \end{array}$$

Further. We have $C_1'(x) = -C_2'(x) \tan(x) = \frac{-\cos(x)}{\sin(x) \times ((\cos(x))^2 - (\sin(x))^2)} \times \tan(x) = \frac{1}{-((\cos(x))^2 - (\sin(x))^2)} = \frac{-1}{\cos(2x)}$. Then, $d(C_1'(x)) = \frac{-dx}{\cos(2x)}$.

$\int d(C_1'(x)) = -\int \frac{dx}{\cos(2x)}$; $C_1(x) = -\int \frac{dx}{\cos(2x)} = -\frac{1}{2} \ln |\tan(x + \frac{\pi}{4})| + C_1$. We get the general solution of the equation: $y = C_1(x) \cos(x) + C_2(x) \sin(x) = -\frac{1}{2} \times \cos(x) \times \ln |\tan(x + \frac{\pi}{4})| + C_1 \cos(x) + \sin(x) \times \ln \left| \frac{\sin(x)}{\sqrt{(1 - 2(\sin(x))^2)}} \right| + C_2 \sin(x)$.

Answer: $y = -\frac{1}{2} \cos(x) \ln |\tan(x + \frac{\pi}{4})| + C_1 \cos(x) + \sin(x) \ln \left| \frac{\sin(x)}{\sqrt{(1 - 2(\sin(x))^2)}} \right| + C_2 \sin(x)$.

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