MAT244--2019F > Term Test 1

Problem 1 (morning)

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Victor Ivrii:
(a) Find integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(-y\sin(x)+y^3\cos(x)\bigr) + \bigl(3\cos(x)+5y^2\sin(x)\bigr) y'=0
\end{equation*}

(b) Also, find a solution satisfying $y(\dfrac{\pi}{4})=\sqrt{2}$.

Ruojing Chen:
(a) Let $$M=-ySin(x)+y^3Cos(x)$$
$$N=3Cos(x)+5y^2Sin(x)$$
Then$$M_y=-Sin(x)+3y^2Cos(x)$$
$$N_x=-3Sinx(x)+5y^2Cos(x)$$
$$R=\frac{M_y-N_x}{M}=\frac{2Sin(x)-2Cos(x)}{-ySin(x)+y^3Cos(x)}=\frac{2(Sin(x)-y^2Cos(x))}{-y(Sin(x)-y^2Cos(x))}=-\frac{2}{y}$$
$$\mu=e^{-\int_Rdy}=e^{\int_\frac{2}{y}dy}=e^{2lny}=e^ln(y^2)=y^2$$

Multiple both side by $$\mu=y^2$$
$$y^2(-ySin(x)+y^3Cos(x))+y^2(3Cos(x)+5y^2Sin(x))=0$$
$$M'=-y^3Sin(x)+y^5Cos(x)$$,$$N'=3y^2Cos(x)+5y^4Sin(x)$$
$$\exists\phi_x,y,such that \phi_x=M',\phi_y=N$$
$$\phi=\int_M'dx=\int_-y^3Sin(x)=y^3Cos(x)+y^5Sin(x)+h(y)$$
$$\phi_y=3y^2Cos(x)+5y^4Sin(x)+h'(y)=N'$$
$$h'(y)=0$$
$$h(y)=c$$

$$\therefore \phi=y^3Cos(x)+y^5Sin(x)=c$$
(b)When $$y(\frac{\pi}{4})=\sqrt{2}$$
$$(\sqrt{2})^3Cos(\frac{\pi}{4})+(\sqrt{2}^5)Sin(\frac{\pi}{4})=2\sqrt{2}*\frac{1}{\sqrt{2}}+(4\sqrt{2}*\frac{1}{\sqrt{2}})=2+4=6$$
$$\therefore c=6$$

$$\phi=y^3Cos(x)+y^5Sin(x)=6$$

What is your real life name? I can find it by email, but I am too lazy  :)

EroSkulled:
Solve :$(-y\sin(x)+y^{3}\cos(x))+(3\cos(x)+5y^{2}\sin(x))y'=0$

M=-y\sin(x)+y^{3}\cos(x), N=3\cos(x)+5y^{2}\sin(x)

M_y=-\sin(x)+3y^{2}\cos(x), N_x=-3\sin(x)+5y^{2}\cos(x)

R_1=\frac{N_x-M_y}{M}=\frac{-3\sin(x)+5y^{2}\cos(x)+\sin(x)-3y^{2}\cos(x)}{-y\sin(x)+y^{3}\cos(x)}=\frac{-2\sin(x)+2y^{2}\cos(x)}{y(-\sin(x)+y^{2}\cos(x))}=\frac{2}{y}

\mu=e^{\int{R_1}{dy}}=e^{\int{\frac{2}{y}}{dy}}=e^{2\ln{y}}=y^2

We then multiply both side of the original equation by $y^2$ so that it will become EXACT and hence we can continue to find $\phi(x,y)$

(-y^{3}\sin(x)+y^{5}\cos(x))+(3y^{2}\cos(x)+5y^{4}\sin(x))y'=0

\phi(x,y)=\int{-y^{3}\sin(x)+y^{5}\cos(x)}{dx}=y^{3}\cos(x)+y^{5}\sin(x)+h(y)

\phi(x,y)_y=3y^{2}\cos(x)+5y^{4}\sin(x)+h'(y)\cong 3y^{2}\cos(x)+5y^{4}\sin(x)

Hence we know $h'(y)=0$
Then $h(y)=C$

\phi(x,y): y^{3}\cos(x)+y^{5}\sin(x)=C

Initial Value: $y(\frac{\pi}{4})=\sqrt{2}$
Plug in equation above, we get the following:

(\sqrt{2})^{3}\cos(\frac{\pi}{4})+(\sqrt{2})^{5}\sin(\frac{\pi}{4})=C

2\sqrt{2}\frac{1}{\sqrt{2}}+4\sqrt{2}\frac{1}{\sqrt{2}}=C

C=6

We get solution:

y^{3}\cos(x)+y^{5}\sin(x)=6

No post after this is needed. V.I.
Instead of sequence single equations it would be better to use multiline environment like gather or gather* to avoid excessive vertical spacing
--- Code: ---\begin{gather} EQUATION \\ EQUATION \\  .... \end{gather}
--- End code ---
If there is no text between them, as MathJax does not support \intertext{  } LaTeX command

EroSkulled:

--- Quote from: rj127 on October 23, 2019, 06:37:18 AM ---(a) Let $$M=-ySin(x)+y^3Cos(x)$$
$$N=3Cos(x)+5y^2Sin(x)$$
Then$$M_y=-Sin(x)+3y^2Cos(x)$$
$$N_x=-3Sinx(x)+5y^2Cos(x)$$
$$R=\frac{M_y-N_x}{M}=\frac{2Sin(x)-2Cos(x)}{-ySin(x)+y^3Cos(x)}=\frac{2(Sin(x)-y^2Cos(x))}{-y(Sin(x)-y^2Cos(x))}=-\frac{2}{y}$$
$$\mu=e^{-\int_Rdy}=e^{\int_\frac{2}{y}dy}=e^{2lny}=e^ln(y^2)=y^2$$

Multiple both side by $$\mu=y^2$$
$$y^2(-ySin(x)+y^3Cos(x))+y^2(3Cos(x)+5y^2Sin(x))=0$$
$$M'=-y^3Sin(x)+y^5Cos(x)$$,$$N'=3y^2Cos(x)+5y^4Sin(x)$$
$$\exists\phi_x,y,such that \phi_x=M',\phi_y=N$$
$$\phi=\int_M'dx=\int_-y^3Sin(x)=y^3Cos(x)+y^5Sin(x)+h(y)$$
$$\phi_y=3y^2Cos(x)+5y^4Sin(x)+h'(y)=N'$$
$$h'(y)=0$$
$$h(y)=c$$

$$\therefore \phi=y^3Cos(x)+y^5Sin(x)=c$$
(b)When $$y(\frac{\pi}{4})=\sqrt{2}$$
$$(\sqrt{2})^3Cos(\frac{\pi}{4})+(\sqrt{2}^5)Sin(\frac{\pi}{4})=2\sqrt{2}*\frac{1}{\sqrt{2}}+(4\sqrt{2}*\frac{1}{\sqrt{2}})=2+4=6$$
$$\therefore c=6$$

$$\phi=y^3Cos(x)+y^5Sin(x)=6$$

--- End quote ---
Above solution is not typed well in correct format so I posted mine as well.

Jiuru Gao:
Miss y^2 of R=My-Nx, it should be 2sinx- 2y^2(cosx)