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nikkih1225

nikkih1225

05.10.2019 •

Mathematics

Solve (1-sinxtany)dx + (cosxsec^2y)dy = 0

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Ответ:

zurfluhe

05.01.2022

This ODE is exact, since

$\dfrac\partial{\partial y}[1-\sin x\tan y]=-\sin x\sec^2y$
$\dfrac\partial{\partial x}[\cos x\sec^2y]=-\sin x\sec^2y$

We're looking for a solution of the form $\Psi(x,y)=C$ such that, by the chain rule,

$\dfrac{\mathrm d}{\mathrm dx}\Psi(x,y)=\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0$
$\implies\begin{cases}\Psi_x=1-\sin x\tan y\\\Psi_y=\cos x\sec^2y\end{cases}$

Integrating the first equation with respect to

, we have

$\displaystyle\int\Psi_x\,\mathrm dx=\int(1-\sin x\tan y)\,\mathrm dx$
$\Psi(x,y)=x+\cos x\tan y+f(y)$

Differentiating with respect to

yields

$\Psi_y=\cos x\sec^2y+f'(y)$
$\cos x\sec^2y=\cos x\sec^2y+f'(y)$
f'(y)=0

f'(y)=0

$\implies f(y)=C$

So the solution to this ODE is

$\Psi(x,y)=x+\cos x\tan y+C=C$

or simply

$\Psi(x,y)=x+\cos x\tan y=C$

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Ответ:

torresnoemi899

26.07.2022

hello! :p

area of shape one is 12

and area of shape to is 36

In not 100% sure but I thought I could try to help! Don't take my word for it though. :) Have a wonderful day!

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