Author Topic: Problem 1 (morning)  (Read 895 times)

Victor Ivrii

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Problem 1 (morning)
« on: October 23, 2019, 05:53:12 AM »
(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

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Re: Problem 1 (morning)
« Reply #1 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 $$y(\frac{\pi}{4})=\sqrt{2}$$
$$(\sqrt{2})^3Cos(\frac{\pi}{4})+(\sqrt{2}^5)Sin(\frac{\pi}{4})=2\sqrt{2}*\frac{1}{\sqrt{2}}+(4\sqrt{2}*\frac{1}{\sqrt{2}})=2+4=6$$
$$\therefore c=6$$

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

What is your real life name? I can find it by email, but I am too lazy  :)
« Last Edit: October 31, 2019, 08:51:58 AM by Victor Ivrii »

EroSkulled

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Re: Problem 1 (morning)
« Reply #2 on: October 23, 2019, 07:27:33 AM »
Solve :$(-y\sin(x)+y^{3}\cos(x))+(3\cos(x)+5y^{2}\sin(x))y'=0$
\begin{equation}
M=-y\sin(x)+y^{3}\cos(x), N=3\cos(x)+5y^{2}\sin(x)
\end{equation}
\begin{equation}
M_y=-\sin(x)+3y^{2}\cos(x), N_x=-3\sin(x)+5y^{2}\cos(x)
\end{equation}
\begin{equation}
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}
\end{equation}
\begin{equation}
\mu=e^{\int{R_1}{dy}}=e^{\int{\frac{2}{y}}{dy}}=e^{2\ln{y}}=y^2
\end{equation}
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 $\phi(x,y)$
\begin{equation}
(-y^{3}\sin(x)+y^{5}\cos(x))+(3y^{2}\cos(x)+5y^{4}\sin(x))y'=0
\end{equation}
\begin{equation}
\phi(x,y)=\int{-y^{3}\sin(x)+y^{5}\cos(x)}{dx}=y^{3}\cos(x)+y^{5}\sin(x)+h(y)
\end{equation}
\begin{equation}
\phi(x,y)_y=3y^{2}\cos(x)+5y^{4}\sin(x)+h'(y)\cong 3y^{2}\cos(x)+5y^{4}\sin(x)
\end{equation}
Hence we know $h'(y)=0$
Then $h(y)=C$
\begin{equation}
\phi(x,y): y^{3}\cos(x)+y^{5}\sin(x)=C
\end{equation}
Initial Value: $y(\frac{\pi}{4})=\sqrt{2}$
Plug in equation above, we get the following:
\begin{equation}
(\sqrt{2})^{3}\cos(\frac{\pi}{4})+(\sqrt{2})^{5}\sin(\frac{\pi}{4})=C
\end{equation}
\begin{equation}
2\sqrt{2}\frac{1}{\sqrt{2}}+4\sqrt{2}\frac{1}{\sqrt{2}}=C
\end{equation}
\begin{equation}
C=6
\end{equation}
We get solution:
\begin{equation}
y^{3}\cos(x)+y^{5}\sin(x)=6
\end{equation}

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: [Select]
\begin{gather} EQUATION \\ EQUATION \\  .... \end{gather} If there is no text between them, as MathJax does not support \intertext{  } LaTeX command
« Last Edit: October 31, 2019, 08:50:24 AM by Victor Ivrii »

EroSkulled

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Re: Problem 1 (morning)
« Reply #3 on: October 23, 2019, 07:30:45 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 $$y(\frac{\pi}{4})=\sqrt{2}$$
$$(\sqrt{2})^3Cos(\frac{\pi}{4})+(\sqrt{2}^5)Sin(\frac{\pi}{4})=2\sqrt{2}*\frac{1}{\sqrt{2}}+(4\sqrt{2}*\frac{1}{\sqrt{2}})=2+4=6$$
$$\therefore c=6$$

$$\phi=y^3Cos(x)+y^5Sin(x)=6$$
Above solution is not typed well in correct format so I posted mine as well.

Jiuru Gao

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Re: Problem 1 (morning)
« Reply #4 on: October 23, 2019, 07:32:53 AM »
Miss y^2 of R=My-Nx, it should be 2sinx- 2y^2(cosx)

Mengyuan Wang

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Re: Problem 1 (morning)
« Reply #5 on: October 23, 2019, 07:42:27 AM »
\begin{equation}
   \left(-y \sin (x)+y^{3} \cos (x)\right)+\left(3 \cos x+5 y^{2} \sin (x)\right) y^{\prime}=0
   \end{equation}
   \begin{equation}
   M _{y}=-\sin (x)+3 y^{2} \cos (x)
   \end{equation}
   \begin{equation}
N_{ x}=-3 \sin (x)+5 y^{2} \cos (x)
   \end{equation}
   \begin{equation}
   M_{y} \neq N_{x}
   \end{equation}
   so not exact
   \begin{equation}
   \begin{aligned} R_{1}=\frac{M_{ y}-N_{ x}}{M} &=\frac{-\sin (x)+3 y^{2} \cos (x)+3 \sin x-5 y^{2} \cos (x)}{-y \sin (x)+y^{3} \cos (x)} \\ &=\frac{2 \sin (x)-2 y^{2} \cos x}{-y \sin (x)+y^{3} \cos (x)}=-\frac{2}{y} \end{aligned}
   \end{equation}
   \begin{equation}
   u=e^{-2\int-\frac{1}{y} d y}=e^{2 \ln y}=y^{2}
   \end{equation}
   so  \begin{equation}
   \left(-y^{3} \sin (x)+y^{5} \cos (x)\right)+\left(3 y^{2} \cos (x)+5 y^{4} \sin (x)\right) y^{\prime}=0
   \end{equation}
   \begin{equation}
   \varphi_{x}=M
   \end{equation}
   \begin{equation}
   \varphi=\int M d x=y^{3} \cos (x)+y^{5} \sin (x)+h(x)
   \end{equation}
   \begin{equation}
   \varphi_{y}=N
   \end{equation}
   \begin{equation}
   \begin{aligned} \varphi _{y }&=3 y^{2} \cos (x)+5 y^{4} \sin (x)+h^{\prime}(x) \\ &=3 y^{2} \cos (x)+5 y^{4} \sin (x) \end{aligned}
   \end{equation}
   so \begin{equation}
   h^{\prime}(x)=0
   \end{equation}
   \begin{equation}
   h(x)=c
   \end{equation}
   so \begin{equation}
   \varphi_{\left(x, y\right)}=y^{3} \cos x+y^{5} \sin x=c
   \end{equation}
   \item because   \begin{equation}
   y\left(\frac{\pi}{4}\right)=\sqrt{2}
   \end{equation}
   plug in
   \begin{equation}
   \begin{array}{c}{(\sqrt{2})^{3} \cos \frac{\pi}{4}+(\sqrt{2})^{5} \sin \frac{\pi}{4}=C} \\ {c=2+4=6}\end{array}
   \end{equation}
   so
   \begin{equation}
   y^{3} \cos x+y^{5} \sin x=6
   \end{equation}

Sally

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Re: Problem 1 (morning)
« Reply #6 on: October 23, 2019, 08:27:05 AM »
(-ysin(x)+(y^3)cos(x))+(3cos(x)+5(y^2)sin(x))y ’=0
a). My=-sin(x) +(3y^2)(cos(x)) Nx = - 3s in ( x ) + 5( y ^ 2) ( c o s( x ) )
R 2= [ My - N x ] /M = [ - si n ( x ) +( 3 y ^2 ) co s ( x ) + 3s in ( x ) - 5( y ^ 2) c o s( x ) ] / [ - y si n ( x ) +( y ^ 3) ( c o s( x ) ) = - 2 / y
u=e^(- ∫(2/y)dy)=y^2 u(-ysin(x)+(y^3)cos(x))+u(3cos(x)+5(y^2)sin(x))y ’=0
Ø=∫u(-ysin(x)+(y^3)cos(x)) dx =(y^3)cos(x)+(y^5)sin(x)
Ø y =3( y ^ 2) c o s( x ) + 5( y ^ 4) s i n ( x )
=3(y^2)cos(x)+5(y^4)sin(x)+h ’(y) h’(y)=0
h(y)=c C=(y^3)cos(x)+(y^5)sin(x)

b) x=π/4 y=√2
C=2+4=6

AllanLi

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Re: Problem 1 (morning)
« Reply #7 on: October 23, 2019, 09:24:54 AM »
\begin{equation}
-ysinx+y^3cosx+(3cosx+5y^2sinx)y'=0
\end{equation}
\begin{equation}
(-ysinx+y^3cosx)dx+(3cosx+5y^2sinx)dy=0
\end{equation}let M = -ysinx+y^3cosx, and N = 3cosx+5y^2sinx
\begin{equation}
M_y = -sinx+3y^2cosx, N_x = -3sinx+5y^2cosx
\end{equation} let My-Nx
\begin{equation}
M_y - N_x = 2sinx-2y^2cosx
\end{equation}since this looks famillier with M , so we are taking R1
\begin{equation}
R_1 = \frac{-M_y+N_x}{M}, R_1 = \frac{-2sinx+2y^2cosx}{-ysinx+y^3cosx},R_1 = \frac{2}{y}
\end{equation} the the integrating factor 𝝻 will be the e to the power of integral of R1
\begin{equation}
𝝻 = e^{∫R_1dy},𝝻 = y^2
\end{equation}then we times y^2 in this equation ,we get
\begin{equation}
(-y^3sinx+y^5cosx)dx+(3y^2cosx+5y^4sinx)dy=0
\end{equation}therefore
\begin{equation}
𝛗 = ∫-y^3sinx+y^5cosxdx=y^3cosx+y^5sinx+h(y)
\end{equation} take the derivative on 𝛗 of y,we get
\begin{equation}
𝛗_y = 3y^2cosx+5y^4sinx+h'(y)=3y^2cosx+5y^4sinx
\end{equation}we will have h'(y) = 0, so h(y)=C is a constant
\begin{equation}
𝛗=y^3cosx+y^5sinx+C
\end{equation}
\begin{equation}
y^3cosx+y^5sinx+C=0
\end{equation}since we have
\begin{equation}
y(\frac{\pi_1}{4})=√2
\end{equation}solve for C
\begin{equation}
C =6
\end{equation}
\begin{equation}
y^3cosx+y^5sinx+6=0
\end{equation}

GuangyuDu

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Re: Problem 1 (morning)
« Reply #8 on: October 23, 2019, 09:57:47 AM »
Question 1:
$
[-y\sin(x)+y^3\cos(x)]+[3\cos(x)+5y^2\sin(x)]y'=0,
y\left(\frac{\pi}{4}\right)=\sqrt2.
$

Solution:
Since $M_y$ does not equal to $N_x$, the equation is not exact.
$
R_1=\frac{M_y-N_x}{M}=\frac{2(\sin x-y^2\cos x)}{-y(\sin x-y^2\cos x)}=-\frac{2}{y}.
$
$
M=e^{-\int R_1\mathrm{d}y}=e^{\int\frac{2}{y}\mathrm{d}y}=e^{2\ln y}=y^2.
$

Multiple both side with $y^2$.
$
[-y^3\sin (x)+y^5\cos(x)]+[3y^2\cos(x)+5y^4\sin(x)]y'=0
$
$
\varphi_x=M
$
$
\varphi=\int M \mathrm{d}x=\int -y^3\sin x +y^5\cos x \mathrm{d}x=y^3\cos x+y^5\sin x +h(y)
$
$
\varphi_y=N=3y^2\cos(x)+5y^4\sin (x)+h'(y)
$
$
h(y)=C
$
$
\varphi (x,y)=y^3\cos x+y^5\sin x=C
$
$
C=\sqrt2^3x\cos \frac{\pi}{4}+\sqrt2^5x\sin \frac{\pi}4=6
$
$
\varphi(x,y)=y^3\cos x+y^5\sin x=6
$
« Last Edit: October 23, 2019, 09:59:28 AM by GuangyuDu »