Answer on Question #81208 – Math – Differential Equations Question
Solve
p 2 + 2 p y cot x = y 2 p^2 + 2py \cot x = y^2 p 2 + 2 p y cot x = y 2
Solution
p 2 + 2 p y cot x = y 2 cot 2 x + 1 = cosec 2 x ⇒ cosec 2 x − cot 2 x = 1 p 2 + 2 p y cot x = y 2 ( cosec 2 x − cot 2 x ) ( p 2 + 2 p y cot x + y 2 cot 2 x ) − y 2 cosec 2 x = 0 ( p + y cot x ) 2 − y 2 cosec 2 x = 0 ( p + y cot x + y cosec x ) ( p + y cot x − y cosec x ) = 0 p + y cot x + y cosec x = 0 or p + y cot x − y cosec x = 0 \begin{array}{l}
p^2 + 2py \cot x = y^2 \\
\cot^2 x + 1 = \cosec^2 x \Rightarrow \cosec^2 x - \cot^2 x = 1 \\
p^2 + 2py \cot x = y^2 (\cosec^2 x - \cot^2 x) \\
(p^2 + 2py \cot x + y^2 \cot^2 x) - y^2 \cosec^2 x = 0 \\
(p + y \cot x)^2 - y^2 \cosec^2 x = 0 \\
(p + y \cot x + y \cosec x)(p + y \cot x - y \cosec x) = 0 \\
p + y \cot x + y \cosec x = 0 \quad \text{or} \quad p + y \cot x - y \cosec x = 0 \\
\end{array} p 2 + 2 p y cot x = y 2 cot 2 x + 1 = cosec 2 x ⇒ cosec 2 x − cot 2 x = 1 p 2 + 2 p y cot x = y 2 ( cosec 2 x − cot 2 x ) ( p 2 + 2 p y cot x + y 2 cot 2 x ) − y 2 cosec 2 x = 0 ( p + y cot x ) 2 − y 2 cosec 2 x = 0 ( p + y cot x + y cosec x ) ( p + y cot x − y cosec x ) = 0 p + y cot x + y cosec x = 0 or p + y cot x − y cosec x = 0 p + y cot x + y cosec x = 0 d y d x + y ( cot x + cosec x ) = 0 d y y = − ( cos x sin x + 1 sin x ) d x d y y = − ( 2 cos 2 ( x 2 ) 2 sin ( x 2 ) cos ( x 2 ) ) d x ∫ d y y = − ∫ cos ( x 2 ) sin ( x 2 ) d x \begin{array}{l}
p + y \cot x + y \cosec x = 0 \\
\frac{dy}{dx} + y (\cot x + \cosec x) = 0 \\
\frac{dy}{y} = -\left(\frac{\cos x}{\sin x} + \frac{1}{\sin x}\right) dx \\
\frac{dy}{y} = -\left(\frac{2 \cos^2 \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}\right) dx \\
\int \frac{dy}{y} = -\int \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} dx \\
\end{array} p + y cot x + y cosec x = 0 d x d y + y ( cot x + cosec x ) = 0 y d y = − ( s i n x c o s x + s i n x 1 ) d x y d y = − ( 2 s i n ( 2 x ) c o s ( 2 x ) 2 c o s 2 ( 2 x ) ) d x ∫ y d y = − ∫ s i n ( 2 x ) c o s ( 2 x ) d x ∫ cos ( x 2 ) sin ( x 2 ) d x \int \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} dx ∫ sin ( 2 x ) cos ( 2 x ) d x
Substitution
u = sin ( x 2 ) , d u = 1 2 cos ( x 2 ) d x ∫ cos ( x 2 ) sin ( x 2 ) d x = 2 ∫ 1 u d u = 2 ln ∣ u ∣ − ln C 1 = 2 ln ∣ sin ( x 2 ) ∣ − ln C 1 ln y = − 2 ln ∣ sin ( x 2 ) ∣ + ln C 1 y = C 1 sin 2 ( x 2 ) = C 1 cosec 2 ( x 2 ) y sin 2 ( x 2 ) = C 1 \begin{array}{l}
u = \sin \left(\frac{x}{2}\right), \quad du = \frac{1}{2} \cos \left(\frac{x}{2}\right) dx \\
\int \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} dx = 2 \int \frac{1}{u} du = 2 \ln |u| - \ln C_1 = 2 \ln |\sin \left(\frac{x}{2}\right)| - \ln C_1 \\
\ln y = -2 \ln |\sin \left(\frac{x}{2}\right)| + \ln C_1 \\
y = \frac{C_1}{\sin^2 \left(\frac{x}{2}\right)} = C_1 \cosec^2 \left(\frac{x}{2}\right) \\
y \sin^2 \left(\frac{x}{2}\right) = C_1 \\
\end{array} u = sin ( 2 x ) , d u = 2 1 cos ( 2 x ) d x ∫ s i n ( 2 x ) c o s ( 2 x ) d x = 2 ∫ u 1 d u = 2 ln ∣ u ∣ − ln C 1 = 2 ln ∣ sin ( 2 x ) ∣ − ln C 1 ln y = − 2 ln ∣ sin ( 2 x ) ∣ + ln C 1 y = s i n 2 ( 2 x ) C 1 = C 1 cosec 2 ( 2 x ) y sin 2 ( 2 x ) = C 1 p + y cot x − y csc x = 0 d y d x + y ( cot x − csc x ) = 0 d y y = ( 1 sin x − cos x sin x ) d x d y y = ( 2 sin 2 ( x 2 ) 2 sin ( x 2 ) cos ( x 2 ) ) d x ∫ d y y = ∫ sin ( x 2 ) cos ( x 2 ) d x \begin{array}{l}
p + y \cot x - y \csc x = 0 \\
\frac{dy}{dx} + y (\cot x - \csc x) = 0 \\
\frac{dy}{y} = \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right) dx \\
\frac{dy}{y} = \left(\frac{2 \sin^{2} \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}\right) dx \\
\int \frac{dy}{y} = \int \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} dx \\
\end{array} p + y cot x − y csc x = 0 d x d y + y ( cot x − csc x ) = 0 y d y = ( s i n x 1 − s i n x c o s x ) d x y d y = ( 2 s i n ( 2 x ) c o s ( 2 x ) 2 s i n 2 ( 2 x ) ) d x ∫ y d y = ∫ c o s ( 2 x ) s i n ( 2 x ) d x ∫ sin ( x 2 ) cos ( x 2 ) d x \int \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} dx ∫ cos ( 2 x ) sin ( 2 x ) d x
Substitution
u = cos ( x 2 ) , d u = − 1 2 sin ( x 2 ) d x ∫ sin ( x 2 ) cos ( x 2 ) d x = − 2 ∫ 1 u d u = − 2 ln ∣ u ∣ + ln C 2 = − 2 ln ∣ cos ( x 2 ) ∣ + ln C 2 ln y = − 2 ln ∣ cos ( x 2 ) ∣ + ln C 2 y = C 2 cos 2 ( x 2 ) = C 2 sec 2 ( x 2 ) y cos 2 ( x 2 ) = C 2 \begin{array}{l}
u = \cos \left(\frac{x}{2}\right), du = -\frac{1}{2} \sin \left(\frac{x}{2}\right) dx \\
\int \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)} dx = -2 \int \frac{1}{u} du = -2 \ln |u| + \ln C_{2} = -2 \ln \left| \cos \left(\frac{x}{2}\right) \right| + \ln C_{2} \\
\ln y = -2 \ln \left| \cos \left(\frac{x}{2}\right) \right| + \ln C_{2} \\
y = \frac{C_{2}}{\cos^{2} \left(\frac{x}{2}\right)} = C_{2} \sec^{2} \left(\frac{x}{2}\right) \\
y \cos^{2} \left(\frac{x}{2}\right) = C_{2} \\
\end{array} u = cos ( 2 x ) , d u = − 2 1 sin ( 2 x ) d x ∫ c o s ( 2 x ) s i n ( 2 x ) d x = − 2 ∫ u 1 d u = − 2 ln ∣ u ∣ + ln C 2 = − 2 ln ∣ ∣ cos ( 2 x ) ∣ ∣ + ln C 2 ln y = − 2 ln ∣ ∣ cos ( 2 x ) ∣ ∣ + ln C 2 y = c o s 2 ( 2 x ) C 2 = C 2 sec 2 ( 2 x ) y cos 2 ( 2 x ) = C 2
The general solution is
( y sin 2 ( x 2 ) − C 1 ) ( y cos 2 ( x 2 ) − C 2 ) = 0. \left(y \sin^{2} \left(\frac{x}{2}\right) - C_{1}\right) \left(y \cos^{2} \left(\frac{x}{2}\right) - C_{2}\right) = 0. ( y sin 2 ( 2 x ) − C 1 ) ( y cos 2 ( 2 x ) − C 2 ) = 0.
Answer: ( y sin 2 ( x 2 ) − C 1 ) ( y cos 2 ( x 2 ) − C 2 ) = 0. \left(y \sin^{2} \left(\frac{x}{2}\right) - C_{1}\right) \left(y \cos^{2} \left(\frac{x}{2}\right) - C_{2}\right) = 0. ( y sin 2 ( 2 x ) − C 1 ) ( y cos 2 ( 2 x ) − C 2 ) = 0.
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#339914 on Dec 2023