Q14Mathematics II

Question

Q.4 Solve : $sin^{2} x \frac{d ^{2} y}{d x ^{2}} - 2 y = 0$

Answer 1

Given the specific second-order linear differential equation strictly with variable coefficients:

sin^{2} x \frac{d ^{2} y}{d x ^{2}} - 2 y = 0 ⟹ \frac{d ^{2} y}{d x ^{2}} - 2 csc^{2} x \cdot y = 0

Step 1: Finding a Known Solution by Inspection

We rigorously test basic trigonometric functions. Let's strictly test $y_{1} = cot x$ :

y_{1}^{'} = - csc^{2} x

y_{1}^{''} = - 2 csc x (- csc x cot x) = 2 csc^{2} x cot x

Substituting exactly back into the given equation:

y_{1}^{''} - 2 csc^{2} x \cdot y_{1} = 2 csc^{2} x cot x - 2 csc^{2} x (cot x) = 0

Thus, $y_{1} = cot x$ is absolutely confirmed as a valid independent solution.

Step 2: Method of Reduction of Order

To find the second linearly independent solution $y_{2}$ , we rigorously utilize the standard formula for reduction of order, given the standard form $y^{''} + P y^{'} + Q y = 0$ (here $P = 0$ ):

y_{2} = y_{1} \int \frac{e ^{- \int P d x}}{y _{1}^{2}} d x

Since $P = 0$ , $e^{- \int P d x} = e^{0} = 1$ .

y_{2} = cot x \int \frac{1}{( cot x ) ^{2}} d x = cot x \int tan^{2} x d x

We systematically apply the trigonometric identity $tan^{2} x = sec^{2} x - 1$ :

y_{2} = cot x \int (sec^{2} x - 1) d x = cot x (tan x - x)

Expanding this strictly yields:

y_{2} = cot x tan x - x cot x = 1 - x cot x

Step 3: Complete General Solution

The absolute complete mathematical solution is the exact linear combination of the two independent solutions:

y = C_{1} y_{1} + C_{2} y_{2} = C_{1} cot x + C_{2} (1 - x cot x)

Q14Mathematics II

Question

Answer

Step 1: Finding a Known Solution by Inspection

Step 2: Method of Reduction of Order

Step 3: Complete General Solution