MAT244--2018F > Final Exam

FE-P1

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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:

--- Quote from: Jingze Wang on December 14, 2018, 09:53:16 AM ---
$\psi(x,y)=x^2 \sin^4(y)+x \sin^3(y)+2\sin^5 (y)$

--- End quote ---

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.