### Author Topic: Problem 1 (morning)  (Read 1957 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 :$(-y\sin(x)+y^{3}\cos(x))+(3\cos(x)+5y^{2}\sin(x))y'=0$

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: [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 »

\left(-y \sin (x)+y^{3} \cos (x)\right)+\left(3 \cos x+5 y^{2} \sin (x)\right) y^{\prime}=0

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

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

M_{y} \neq N_{x}

so not exact

\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}

u=e^{-2\int-\frac{1}{y} d y}=e^{2 \ln y}=y^{2}

so
\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

\varphi_{x}=M

\varphi=\int M d x=y^{3} \cos (x)+y^{5} \sin (x)+h(x)

\varphi_{y}=N

\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}

so
h^{\prime}(x)=0

h(x)=c

so
\varphi_{\left(x, y\right)}=y^{3} \cos x+y^{5} \sin x=c

\item because
y\left(\frac{\pi}{4}\right)=\sqrt{2}

plug in

\begin{array}{c}{(\sqrt{2})^{3} \cos \frac{\pi}{4}+(\sqrt{2})^{5} \sin \frac{\pi}{4}=C} \\ {c=2+4=6}\end{array}

so

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

#### 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 »

-ysinx+y^3cosx+(3cosx+5y^2sinx)y'=0

(-ysinx+y^3cosx)dx+(3cosx+5y^2sinx)dy=0
let M = -ysinx+y^3cosx, and N = 3cosx+5y^2sinx

M_y = -sinx+3y^2cosx, N_x = -3sinx+5y^2cosx
let My-Nx

M_y - N_x = 2sinx-2y^2cosx
since this looks famillier with M , so we are taking R1

R_1 = \frac{-M_y+N_x}{M}, R_1 = \frac{-2sinx+2y^2cosx}{-ysinx+y^3cosx},R_1 = \frac{2}{y}
the the integrating factor 𝝻 will be the e to the power of integral of R1

𝝻 = e^{∫R_1dy},𝝻 = y^2
then we times y^2 in this equation ,we get

(-y^3sinx+y^5cosx)dx+(3y^2cosx+5y^4sinx)dy=0
therefore

𝛗 = ∫-y^3sinx+y^5cosxdx=y^3cosx+y^5sinx+h(y)
take the derivative on 𝛗 of y,we get

𝛗_y = 3y^2cosx+5y^4sinx+h'(y)=3y^2cosx+5y^4sinx
we will have h'(y) = 0, so h(y)=C is a constant

𝛗=y^3cosx+y^5sinx+C

y^3cosx+y^5sinx+C=0
since we have

y(\frac{\pi_1}{4})=√2
solve for C

C =6

y^3cosx+y^5sinx+6=0

#### 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 »