FE-P1

[Victor Ivrii]

Typed solutions only. No uploads

Find the general solution
\begin{equation*}
\bigl[2x\sin(y) +1\bigr]\,dx  +
\bigl[4x^2\cos(y) + 3x\cot(y)+5 \sin(2y)\bigr]\,dy=0\,.
\end{equation*}
Hint: Use the integrating factor.

[Jingze Wang]

Let $M=2x\sin(y)+1, N=4x^2\cos(y)+3x\cot(y)+5\sin(2y)$
$M_y=2x\cos(y), N_x=8x\cos(y)+3\cot(y)$
Check and find this is not exact
Then try to find integrating factor
$N_x-M_y=8x\cos(y)+3\cot(y)-2x\cos(y)=6x\cos(y)+3\cot(y)$
By observation, $\frac{N_x-M_y}{M}=3\cot(y)$
Therefore, the integrating factor is $\sin^{3}(y)$
$M'=2x\sin^4(y)+\sin^3(y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+h(y)$
$\psi_y=4x^2\sin^3(y)\cos(y) + x \sin^2(y)\cos(y)+h'(y)$
$h'(y)=10\sin^4(y) \cos(y)$
$h(y)=2\sin^5 (y)$
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$

[Jingyi Wang]

Do not forget the function of Y. We also have to integrate N.
The final answer should be:
𝑥²sin⁴𝑦+𝑥sin³𝑦+2sin⁵y

[Jingze Wang]

Sorry I have not finished my typed solution at that time, so you just saw part of my solution, but we get the same answer finally

[Jingyi Wang]

Sorry about that. I did not see that part. Please ignore my post.

[Tianfangtong Zhang]

$M = 2x \sin(y) +1 $

$N = 4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y) $

$M_y = 2x\cos(y)$

$N_x = 8x\cos(y) + 3cos(y)$

$\frac{M_y - N_x} {M} = \frac{-6x\cos(y) - 3cos(y)}{2x\sin(y) + 1} = -3\cot(y)$

$\mu = e^{-\int -3\cot(y) dy } = \sin^3(y)$

Then multiply $sin^3(y)$ on M and N.

Then we get $M =  sin^3(y)(2x \sin(y) +1)$, $N =sin^3(y)(4 x^2 \cos(y) + 3x \cot(y) + 5 \sin(2y))$

$M_y = \sin^2(y) \cos(y) (8x \sin(y) + 3) $, and $N_x = \sin^2(y) \cos(y) (8x \sin(y) + 3) $

Thus $M_y = N_x$, exact.

Then $\psi_x = M$

$\psi = \int M dx = \int \sin^3(y)(2x \sin(y) +1) dx = \sin^3(y)(x^2\sin(y) + x) + h(y) $

$N = \psi_y = 4x^2\sin^3(y)\cos(y) + x\sin^2(y)\cos(y) + h'(y) $

$h'(y) = 10\sin^4cos(y) $

$h(y) = \int 10\sin^4cos(y) dy = 2\sin^5(y) + c$

Then $\psi = \sin^3(y)(x^2\sin(y) + x) + 2\sin^5(y) = c$

[Qinger Zhang]

$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$

there should be a 'c', so we get $\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$+c
Does not matter where we put it--to the right or the left. V.I.

[Victor Ivrii]

As $M=2x\sin(y) +1$, $N= 4x^2\cos(y) + 3x\cot(y)+10 \sin(y)\cos(y)$ we have
$$
M_y-N_x= 2x\cos(y)-[8x\cos(y)+3\cot(y)]=-6x \cos(y) -3\frac{\cos(y)}{\sin(y)}
\implies
\frac{N_x-M_y}{M}=  3\cot(y).
$$
Therefore we are looking for an integrating factor $\mu(y)$, satisfying
$$
\frac{\mu'}{\mu}= 3\cot(y)\implies \ln (\mu)=3\int \cot(y)\,dy=3\ln (\sin(y))
$$
(we take constant equal $0$). Then $\mu =\sin^3(y)$.

After multiplication we get
$$
\bigl[2x\sin^4(y) +\sin^3(y)\bigr]\,dx  +
\bigl[4x^2\sin^3(y)\cos(y) + 3x\sin^2(y)\cos(y)+10 \sin^4(y)\cos(y)\bigr]\,dy=0\,.
$$
Looking for $H(x,y)$ satisfying
\begin{align}
&H_x = 2x\sin^4(y) +\sin^3(y)\,,\label{eq-1-1}\\
&H_y=4x^2\sin^3(y)\cos(y) + 3x\sin^2(y)\cos(y)+10 \sin^4(y)\cos(y)\,.
\label{eq-1-2}

\end{align}
(\ref{eq-1-1}) implies
$$H(x,y)= \int \bigl[2x\sin^4(y) +\sin^3(y)\bigr]\,dx=
x^2\sin^4(y)+ x\sin^3(y) +h(y).$$
Then
$$
H_y= 4x^2\sin^3(y)\cos(y)+ 3x\sin^2\cos(y) +h'(y)
$$
and comparing with (\ref{eq-1-2}) we see that
$$
h'(y)=10\sin^4(y)\cos(y)\implies h(y)=\int 10\sin^4(y)\cos(y)\,dy= 2\sin^5(y);
$$
we pick-up constant $0$.


Finally
$$
H(x,y)=
x^2\sin^4(y)+ x\sin^3(y) +2\sin^5(y) =C
$$
is a solution to the original problem.