MAT244--2019F > Term Test 1

Problem 1 (morning)

<< < (2/2)

Mengyuan Wang:

\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:
(-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:

-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:
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 $y^2$.
$[-y^3\sin (x)+y^5\cos(x)]+[3y^2\cos(x)+5y^4\sin(x)]y'=0$
$\varphi_x=M$
$\varphi=\int M \mathrm{d}x=\int -y^3\sin x +y^5\cos x \mathrm{d}x=y^3\cos x+y^5\sin x +h(y)$
$\varphi_y=N=3y^2\cos(x)+5y^4\sin (x)+h'(y)$
$h(y)=C$
$\varphi (x,y)=y^3\cos x+y^5\sin x=C$
$C=\sqrt2^3x\cos \frac{\pi}{4}+\sqrt2^5x\sin \frac{\pi}4=6$
$\varphi(x,y)=y^3\cos x+y^5\sin x=6$