Find:d2y/dx2 + y = cot x\(\frac{d^2y}{dx^2}+y = cotx\)

1.	Find:d2y/dx2 + y = cot x\(\frac{d^2y}{dx^2}+y = cotx\)
Answer» \(\frac{d^2y}{dx^2}+y = cotx\) \(\therefore\) It's complementary equation is m² + m = 0 ⇒ m(m + 1) = 0 ⇒ m = 0, -1 C. F. = Ge^0x + C₂e^-x = C₁ + C₂e^-x Let y₁ = 1, y₂ = e^-x w(y_1, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}1&e^{-x}\\0&-e^{-x}\end{vmatrix}\) = -cot x e^-x Q(x) = cot x w₁ = \(\begin{vmatrix}0&y_2\\Q(x)&y'_2\end{vmatrix}=\) \(\begin{vmatrix}0&e^{-x}\\cot x&-e^{-x}\end{vmatrix}=-cot x e^{-x}\) and w₂ = \(\begin{vmatrix}y_1&0\\y'_1&Q(x)\end{vmatrix}\)\(=\begin{vmatrix}1&0\\0&cotx\end{vmatrix}\) = cot x Now, u₁ = \(\int\frac{w_1}{w}dx=\int\frac{-cotxe^{-x}}{-e^{-x}}dx\) = \(\int cot xdx=log sin\,x\) and u₂ = \(\int\frac{w_2}{w}dx = \int\frac{cotx}{-e^{-x}}dx\) \(=-\int e^xcot\,xdx\) \(\therefore\) P.I. = u₁y₁ + u₂y₂ = log sin x - e^-x\(\int e^xcot\,x dx\) \(\therefore\) Complete solution is y = C.F. + P. I. = C₁ + C₂ e^-x + log sinx - e^-x\(\int e^xcot\,xdx\)

Answer»

\(\frac{d^2y}{dx^2}+y = cotx\)

\(\therefore\) It's complementary equation is

m² + m = 0

⇒ m(m + 1) = 0

⇒ m = 0, -1

C. F. = Ge^0x + C₂e^-x = C₁ + C₂e^-x

Let y₁ = 1, y₂ = e^-x

w(y_1, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}1&e^{-x}\\0&-e^{-x}\end{vmatrix}\) = -cot x e^-x

Q(x) = cot x

w₁ = \(\begin{vmatrix}0&y_2\\Q(x)&y'_2\end{vmatrix}=\) \(\begin{vmatrix}0&e^{-x}\\cot x&-e^{-x}\end{vmatrix}=-cot x e^{-x}\)

and w₂ = \(\begin{vmatrix}y_1&0\\y'_1&Q(x)\end{vmatrix}\)\(=\begin{vmatrix}1&0\\0&cotx\end{vmatrix}\) = cot x

Now, u₁ = \(\int\frac{w_1}{w}dx=\int\frac{-cotxe^{-x}}{-e^{-x}}dx\)

= \(\int cot xdx=log sin\,x\)

and u₂ = \(\int\frac{w_2}{w}dx = \int\frac{cotx}{-e^{-x}}dx\)

\(=-\int e^xcot\,xdx\)

\(\therefore\) P.I. = u₁y₁ + u₂y₂

= log sin x - e^-x\(\int e^xcot\,x dx\)

\(\therefore\) Complete solution is

y = C.F. + P. I.

= C₁ + C₂ e^-x + log sinx - e^-x\(\int e^xcot\,xdx\)

Discussion