Answer on Question #71135 – Math – Differential Equations
Question
Solve the partial differential equation p cos ( x + y ) + q sin ( x + y ) = z p \cos(x+y) + q \sin(x+y)=z p cos ( x + y ) + q sin ( x + y ) = z z z z x , y x,y x , y p p p q q q
Solution
∂ z ∂ x cos ( x + y ) + ∂ z ∂ y sin ( x + y ) = z \frac{\partial z}{\partial x} \cos(x + y) + \frac{\partial z}{\partial y} \sin(x + y) = z ∂ x ∂ z cos ( x + y ) + ∂ y ∂ z sin ( x + y ) = z d x cos ( x + y ) = d y sin ( x + y ) = d z z \frac{dx}{\cos(x + y)} = \frac{dy}{\sin(x + y)} = \frac{dz}{z} cos ( x + y ) d x = sin ( x + y ) d y = z d z { d y sin ( x + y ) = d z z d x cos ( x + y ) = d z z \left\{ \begin{array}{l} \frac{dy}{\sin(x + y)} = \frac{dz}{z} \\ \frac{dx}{\cos(x + y)} = \frac{dz}{z} \end{array} \right. { s i n ( x + y ) d y = z d z c o s ( x + y ) d x = z d z
Solve the first equation of the system:
∫ d y sin ( x + y ) = ∫ d z z \int \frac{dy}{\sin(x + y)} = \int \frac{dz}{z} ∫ sin ( x + y ) d y = ∫ z d z ∫ d y sin ( x + y ) = ∫ sin ( x + y ) 1 − ( cos ( x + y ) ) 2 d x = { let u = cos ( x + y ) and d u = − sin ( x + y ) d x } = \int \frac{dy}{\sin(x + y)} = \int \frac{\sin(x + y)}{1 - (\cos(x + y))^2} dx = \{ \text{let } u = \cos(x + y) \text{ and } du = -\sin(x + y) dx \} = ∫ sin ( x + y ) d y = ∫ 1 − ( cos ( x + y ) ) 2 sin ( x + y ) d x = { let u = cos ( x + y ) and d u = − sin ( x + y ) d x } = ∫ − 1 1 + u 2 d u = − 1 2 ∫ 1 1 + u d u − 1 2 ∫ 1 1 − u d u = ln ( u − 1 ) 2 − ln ( u + 1 ) 2 + Const = ln ( cos ( x + y ) − 1 ) 2 − \int \frac{-1}{1 + u^2} du = \frac{-1}{2} \int \frac{1}{1 + u} du - \frac{1}{2} \int \frac{1}{1 - u} du = \frac{\ln(u - 1)}{2} - \frac{\ln(u + 1)}{2} + \text{Const} = \frac{\ln(\cos(x + y) - 1)}{2} - ∫ 1 + u 2 − 1 d u = 2 − 1 ∫ 1 + u 1 d u − 2 1 ∫ 1 − u 1 d u = 2 ln ( u − 1 ) − 2 ln ( u + 1 ) + Const = 2 ln ( cos ( x + y ) − 1 ) − − ln ( cos ( x + y ) + 1 ) 2 + Const , - \frac{\ln(\cos(x + y) + 1)}{2} + \text{Const}, − 2 ln ( cos ( x + y ) + 1 ) + Const ,
where Const \text{Const} Const
ln ( cos ( x + y ) − 1 ) 2 − ln ( cos ( x + y ) + 1 ) 2 + Const = ln ∣ z ∣ \frac{\ln(\cos(x + y) - 1)}{2} - \frac{\ln(\cos(x + y) + 1)}{2} + \text{Const} = \ln|z| 2 ln ( cos ( x + y ) − 1 ) − 2 ln ( cos ( x + y ) + 1 ) + Const = ln ∣ z ∣ z = ± cos ( x + y ) − 1 cos ( x + y ) + 1 e Const z = \pm \frac{\sqrt{\cos(x + y) - 1}}{\sqrt{\cos(x + y) + 1}} e^{\text{Const}} z = ± cos ( x + y ) + 1 cos ( x + y ) − 1 e Const z = C 1 cos ( x + y ) − 1 cos ( x + y ) + 1 z = C1 \sqrt{\frac{\cos(x + y) - 1}{\cos(x + y) + 1}} z = C 1 cos ( x + y ) + 1 cos ( x + y ) − 1
Solve the second equation of the system:
∫ d x cos ( x + y ) = ∫ d z z \int \frac {dx}{\cos(x + y)} = \int \frac {dz}{z} ∫ cos ( x + y ) d x = ∫ z d z ∫ d y cos ( x + y ) = ∫ cos ( x + y ) 1 − ( sin ( x + y ) ) 2 d x = { l e t u = sin ( x + y ) a n d d u = cos ( x + y ) d x } = ∫ 1 1 − u 2 d u = \begin{aligned}
\int \frac {dy}{\cos(x + y)} &= \int \frac {\cos(x + y)}{1 - (\sin(x + y))^{2}} dx = \{let \, u = \sin(x + y) \, and \, du = \cos(x + y) \, dx\} \\
&= \int \frac {1}{1 - u^{2}} du =
\end{aligned} ∫ cos ( x + y ) d y = ∫ 1 − ( sin ( x + y ) ) 2 cos ( x + y ) d x = { l e t u = sin ( x + y ) an d d u = cos ( x + y ) d x } = ∫ 1 − u 2 1 d u = = 1 2 ∫ 1 1 + u d u + 1 2 ∫ 1 1 − u d u = ln ( u + 1 ) 2 − ln ( u − 1 ) 2 + C o n s t = ln ( sin ( x + y ) + 1 ) 2 − ln ( sin ( x + y ) − 1 ) 2 + C o n s t , = \frac {1}{2} \int \frac {1}{1 + u} du + \frac {1}{2} \int \frac {1}{1 - u} du = \frac {\ln(u + 1)}{2} - \frac {\ln(u - 1)}{2} + Const = \frac {\ln(\sin(x + y) + 1)}{2} - \frac {\ln(\sin(x + y) - 1)}{2} + Const, = 2 1 ∫ 1 + u 1 d u + 2 1 ∫ 1 − u 1 d u = 2 ln ( u + 1 ) − 2 ln ( u − 1 ) + C o n s t = 2 ln ( sin ( x + y ) + 1 ) − 2 ln ( sin ( x + y ) − 1 ) + C o n s t ,
where Const is an integration constant.
ln ( sin ( x + y ) + 1 ) 2 − ln ( sin ( x + y ) − 1 ) 2 + C o n s t = ln ∣ z ∣ \frac {\ln(\sin(x + y) + 1)}{2} - \frac {\ln(\sin(x + y) - 1)}{2} + Const = \ln |z| 2 ln ( sin ( x + y ) + 1 ) − 2 ln ( sin ( x + y ) − 1 ) + C o n s t = ln ∣ z ∣ z = ± sin ( x + y ) + 1 sin ( x + y ) − 1 e C o n s t z = \pm \frac {\sqrt {\sin(x + y) + 1}}{\sqrt {\sin(x + y) - 1}} e^{Const} z = ± sin ( x + y ) − 1 sin ( x + y ) + 1 e C o n s t z = C 2 sin ( x + y ) + 1 sin ( x + y ) − 1 z = C2 \sqrt {\frac {\sin(x + y) + 1}{\sin(x + y) - 1}} z = C 2 sin ( x + y ) − 1 sin ( x + y ) + 1
Then
{ C 1 = z cos ( x + y ) + 1 cos ( x + y ) − 1 C 2 = z sin ( x + y ) − 1 sin ( x + y ) + 1 \left\{
\begin{array}{l}
C1 = z \sqrt {\frac {\cos(x + y) + 1}{\cos(x + y) - 1}} \\
C2 = z \sqrt {\frac {\sin(x + y) - 1}{\sin(x + y) + 1}}
\end{array}
\right. ⎩ ⎨ ⎧ C 1 = z c o s ( x + y ) − 1 c o s ( x + y ) + 1 C 2 = z s i n ( x + y ) + 1 s i n ( x + y ) − 1
The solution of the partial differential equation is given by
F ( z cos ( x + y ) + 1 cos ( x + y ) − 1 ; z sin ( x + y ) − 1 sin ( x + y ) + 1 ) = 0 , F \left( z \sqrt {\frac {\cos(x + y) + 1}{\cos(x + y) - 1}}; z \sqrt {\frac {\sin(x + y) - 1}{\sin(x + y) + 1}} \right) = 0, F ( z cos ( x + y ) − 1 cos ( x + y ) + 1 ; z sin ( x + y ) + 1 sin ( x + y ) − 1 ) = 0 ,
where F F F
Answer:
F ( z cos ( x + y ) + 1 cos ( x + y ) − 1 ; z sin ( x + y ) − 1 sin ( x + y ) + 1 ) = 0 , F \left( z \sqrt {\frac {\cos(x + y) + 1}{\cos(x + y) - 1}}; z \sqrt {\frac {\sin(x + y) - 1}{\sin(x + y) + 1}} \right) = 0, F ( z cos ( x + y ) − 1 cos ( x + y ) + 1 ; z sin ( x + y ) + 1 sin ( x + y ) − 1 ) = 0 ,
where F F F
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#340153 on Dec 2023