Tag Archives: Ordinary Differential Equations

Elementary Differential Equations by Dela Fuente, Feliciano and Uy Physical Application 2: Exponential Growth and Decay


A certain radioactive material follows the law of exponential change and has a half life of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.


Solution:

Use the formula:

S=CektS=Ce^{-kt}

First, find the constant of proportionality. In the problem, after 38 hours, half of the radioactivity has been dissipated and a half has been retained. So we can assume that S = 0.5So when t = 38 hrs and C = So.

(0.5)So=(So)ek(38)\left(0.5\right)So=\left(So\right)e^{-k\left(38\right)}

And then solve for k:

k=0.018241k=-0.018241

And then substitute k to the formula:

S=Ce0.018241(t)S=Ce^{-0.018241\left(t\right)}

Now we can solve for the time(t). According to the problem, 90% of the radioactivity is dissipated, so 10% is retained. So we can assume that S = 0.1So and change C = So.

(0.1)So=(So)e0.018241(t)\left(0.1\right)So=\left(So\right)e^{-0.018241\left(t\right)}

And then solve for time(t):

t=126.23hrst\:=\:126.23\:hrs

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Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 10 Problem 3 — Applications of Ordinary First-Ordered Differential Equations


A tank contains 400 liters of brine. Twelve liters of brine, each containing 2.5 N of dissolved salt, enter the tank per minute, and the mixture, assumed uniform leaves at the rate of 8 liters per min. If the concentration is to be 2 N/litre at the end of one hour, how many newtons of salt should there be present in the tank originally?


Solution:

Consider the following illustration

dSdt=(dSdt)en(dSdt)es\frac{dS}{dt}=\left(\frac{dS}{dt}\right)_{en}-\left(\frac{dS}{dt}\right)_{es}

Using,

Vbrine+(rateofbrineout)tdSdt=8LM(S400+4t)=8S(400+4t)dSdt=2S(100+t)V_{brine}+\left(rate\:of\:brine\:out\right)t\\\frac{dS}{dt}=\frac{8L}{M}\left(\frac{S}{400+4t}\right)\\=\frac{8S}{\left(400+4t\right)}\\\frac{dS}{dt}=\frac{2S}{\left(100+t\right)}

Using the general solution:

dSdt=302S(100+t)dSdt+2S(100+t)=30\frac{dS}{dt}=30-\frac{2S}{\left(100+t\right)}\\\frac{dS}{dt}+\frac{2S}{\left(100+t\right)}=30

To solve we will use First Order Linear Differential Equation (FOLDE) where:

P(t)=2(100+t),Q(t)=30P_{\left(t\right)}=\frac{2}{\left(100+t\right)}\:,\:Q_{\left(t\right)}=30

Solve for the integrating factor using the formula:

σ=eP(t)dt\sigma =e^{\int \:P_{\left(t\right)}dt}

Apply,

σ=e2100+tdtσ=e2ln(100+t)σ=eln(100+t)2σ=(100+t)2\sigma =e^{\int \:\frac{2}{100+t}dt}\\\sigma =e^{2ln\left(100+t\right)}\\\sigma \:=e^{ln\left(100+t\right)^2}\\\sigma \:=\left(100+t\right)^2

Substitute the given value to the formula:

Sσ=σQ(t)dt+CS\sigma =\int \:\sigma Q\left(t\right)dt+C

Apply,

S(100+t)2=(100+t)230dt+CS(100+t)2=30(100+t)2dt+CS(100+t)2=30(100+t)33dt+CS(100+t)2=10(100+t)3+Ceqn.1S\left(100+t\right)^2=\int \:\left(100+t\right)^230dt+C\\S\left(100+t\right)^2=30\int \:\left(100+t\right)^2dt+C\\S\left(100+t\right)^2=30\:\frac{\left(100+t\right)^{^3}}{3}dt+C\\S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C\rightarrow eqn.1

Evaluate C; @t=1hr

Convert 1hr to minutes, where 1hr is simply 60 minutes.

S(100+60)2=10(100+60)3C=2NL;C=S(400+4t)S\left(100+60\right)\:2\:=10\left(100+60\right)\:^3\\C=\frac{2N}{L}\:;\:C=\frac{S}{\left(400+4t\right)}

Get the value of S using the equation:

C=S(400+4t)\:C=\frac{S}{\left(400+4t\right)}

Isolate S,

S=C(400+4t);C=2,t=60S=2(400+4(60))S=1280NS=C\left(400+4t\right);\:C=2,\:t=60\\S=2\left(400+4\left(60\right)\right)\\S=1280N

Get the value of C using Eqn.1

S(100+t)2=10(100+t)3+C;S=1280,t=601280(100+60)2=10(100+60)3+C32768000=40960000+C3276800040960000=CC=8192000S\left(100+t\right)^2=10\left(100+t\right)^{^3}+C; S=1280 , t=60\\ 1280\left(100+60\right)^2=10\left(100+60\right)^{^3}+C\\32768000=40960000+C\\32768000-40960000=C\\ C=-8192000

With the presence of the value of C we will now have our working equation:

S(100+t)2=10(100+t)38192000S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000

Using the given working equation, solve for the value of S @ t=0

S(100+t)2=10(100+t)38192000;t=0S(100+0)2=10(100+0)38192000S(1000)21000=18080001000S=180.8NS\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000; t=0\\S\left(100+0\right)^2=10\left(100+0\right)^{^3}-8192000\\\frac{S\left(1000\right)^2}{1000}=\frac{1808000}{1000}\\S=180.8 N

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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 10 Problem 2 — Applications of Ordinary First-Ordered Differential Equations


Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.


Solution:

Plot points on the curve,

A(x1,y1)A(x_{1},y_{1})

We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.

yy1=1m(xx1)y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)

As the normal intersects the x-axis, y = 0

Substituting to the previous equation, we get

y1=1m(xx1)my1=1(xx1)my1=x+x1x=x1+my1\begin{align*} -y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\ -my_{1}&=-1\left(x-x_{1}\right)\\ -my_{1}&=-x+x_{1}\\ x&=x_{1}+my_{1} \end{align*}

By using distance formula, from the origin (0,0), to point (x1+y1) = from intersection to (x1+y1)

(x10)2+(y10)2=(x1+my1x1)2+(0y1)2x12+y12=m2y12+y12x12+y12=m2y12+y12x12=m2y12x1=m1y1    ;m=dydxx1=dydxy1\begin{align*} \sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\ \sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\ x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\ x_{1}^2&=m^2y_{1}^2\\ x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\ x_{1}&=\frac{dy}{dx}y_{1} \end{align*}

Change x1 and y1 to x and y,

x=ydydxxdx=ydy\begin{align*} x&=y\frac{dy}{dx}\\ xdx&=ydy \end{align*}

By integrating,

ydx=xdyy22=x22+Cy2=x2+2C\begin{align*} \int \:ydx&=\int \:xdy\\ \frac{y^2}{2}&=\frac{x^2}{2}+C\\ y^2&={x^2}+2C\\ \end{align*}

We get,

y2x2=2Cy^2-x^2-=2C

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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 1 – Special Second-Ordered Differential Equations


Find the General Solution

ddx(dydx)=6x+3\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3

Solution:

ddx(dydx)=6x+3solve  the  equation  using  case  1,let  u=dydxdudx=(6x+3)using  separation  of  variable  divide  both  sides  by  dx,du=(6x+3)dxby  integrating  using  the  sum  rule:we  get,u=3x2+3x+C1substitute  the  value  of  u=dydxdydx=3x2+3x+C1using  separation  of  variable:dy=(3x2+3x+C1)dxapply  the  sum  rule:dy=3x2dx+3xdx+C1dxy=x3+3x22+C1x+C2\frac{d}{dx}\left(\frac{dy}{dx}\right)=\:6x\:+\:3 \\ solve\; the\; equation\; using\; case\; 1,\\ let\; u=\frac{dy}{dx} \\ \int \:\frac{du}{dx}\:=\:\int \:\left(6x\:+\:3\right)\\ using\; separation\; of\; variable\; divide\; both\; sides\; by\; dx,\\ \int \:du\:=\:\int \:\left(6x\:+\:3\right)dx\\ by\; integrating\; using\; the\; sum\; rule:\\ we\; get,\\ u=3x^2\:+\:3x\:+\:C_1\\ substitute\; the\; value\; of\; u=\frac{dy}{dx} \\ \frac{dy}{dx}=3x^2\:+\:3x\:+\:C_1\\ using\; separation\; of\; variable:\\ \int \:dy = \int \:\left(3x^2+3x\:+\:C_1\right)dx\\ apply\; the\; sum\; rule:\\ \int \:dy=\int \:3x^2dx+\int \:3xdx+ \int \:C_1dx\\ y=x^3+\frac{3x^2}{2}+C_1x+C_2\\

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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 4 — Special Second-Ordered Differential Equations


Find the general solution of the differential equation

y+2y6x3=0y''+2y'-6x-3=0

Solution:

A second-order differential equation can be written in the form:

ay+by+cy=g(x)ay″ + by' + cy = g(x)

Therefore, the problem given is a second order linear equation.

Here are STEPS on how to get the general solution:

i. simplify the equation to a Second ODE Form

y+2y=6x+3y''+2y'=6x+3

ii. Let

y=P=dydxandy=dPdxdPdx+P2=6x+3\begin{align*} y' = P=\frac{dy}{dx} \\ \\ and \\ \\ y'' = \frac{dP}{dx} \\ \\ \frac{dP}{dx}+P2\:=\:6x+3 \end{align*}

iii. By recalling, we can see that the equation is in First-order linear differential equation form. Solving the simplified equation using FOLDE.

P(x)=2andQ(x)=6x+3dPdx+P2=6x+3\begin{align*} P(x) & =2 \\ and \\ Q(x) &=6x+3 \\ \\ \frac{dP}{dx}+P2\:& =\:6x+3 \end{align*}

Find the integrating factor

ɸ=eP(x)dxɸ=e  2dxɸ=e2x\begin{align*} ɸ & =e^{\int \:P\left(x\right)dx} \\ ɸ & =e^{\int \:\:2dx} \\ ɸ & =e^{2x} \end{align*}

Substituting the I.F. to the formula

Pɸ=ɸQ(x)dx+C1Pe2x=e2x(6x+3)dx+C1Pe2x=(e2x6x+3e2x)dx+C1\begin{align*} Pɸ & =\int \:ɸQ\left(x\right)dx+C_1 \\ Pe^{2x} & =\int \:e^{2x}(6x+3)dx+C_1 \\ Pe^{2x} & =\int \:\left(e^{2x}6x+3e^{2x}\right)dx+C_1 \end{align*}

Integrating the first term

e2x6xdx=6xe2xdx\begin{align*} \int \:e^{2x}6xdx= 6\cdot \int \:xe^{2x}dx \\ \end{align*}

Let u = 2x and du/2 = dx

32euudu\frac{3}{2}\int \:e^uudu

By IBP, Let v=u, dv=du and eudu, n=eu.

nvndvueu32  eudu=euueu=3e2xx32e2x\begin{align*} nv-\int ndv & \\ ue^u-\frac{3}{2}\int \:\:e^udu & = e^uu-e^u \\ & =3e^{2x}x-\frac{3}{2}e^{2x} \end{align*}

for the second term

3e2xdx\int \:3e^{2x}dx

Let u = 2x and du/2 = dx

32eudu=32eu=32e2x\begin{align*} \frac{3}{2}\int \:e^udu & =\frac{3}{2}e^u \\ & =\frac{3}{2}e^{2x} \end{align*}

Combining all the solved terms we get

Pe2x=3e2xx32e2x+32e2x+C1Pe2x=3e2xx+C1\begin{align*} Pe^{2x} & =3e^{2x}x-\frac{3}{2}e^{2x}+\frac{3}{2}e^{2x}+C_1 \\ Pe^{2x} & =3e^{2x}x+C_1 \end{align*}

Based on the equation that we derived it is now a separable differential equation, therefore,

[dydxe2x=3e2xx+C1]1e2xdydx=3x+C1e2xdy=(3x+C1e2x)dx+C2\begin{align*} &\left[\frac{dy}{dx}e^{2x}=3e^{2x}x+C_1\right]\frac{1}{e^{2x}} \\ \frac{dy}{dx} & =3x+\frac{C_1}{e^{2x}} \\ \int \:dy & =\int \:\left(3x+\frac{C_1}{e^{2x}}\right)dx+C_2 \end{align*}

GENERAL SOLUTION:

y=3x22C1e2x2+C2y=\frac{3x^2}{2}-\frac{C_1e^{-2x}}{2}+C_2

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Elementary Differential Equations by Dela Fuente, Feliciano and Uy Chapter 9 Problem 5 — Special Second-Ordered Differential Equations


Find the general solution of the differential equation

yy+2(y)2=0yy''+2\left(y\right)^2=0

Solution:

Based on Special Second-Ordered Differential Equation: Special case 3

F(d2ydx2,dydx,y)=0F\left(\frac{d^2y}{dx^2},\:\frac{dy}{dx},\:y\right)=0

Denote and substitute to the given equation.

P=y=dydxPdpdy=y=d2ydx2P= y' =\frac{dy}{dx} \\ P\frac{dp}{dy}= y'' =\frac{d^2y}{dx^2}

We will have,

y(Pdpdx)+2(P)2=0y(P\frac{dp}{dx})+2(P)^2=0

Divide both sides with

1yP \:\frac{1}{yP}

We will come to,

dpdy+2Py=0\frac{dp}{dy}+\frac{2P}{y}=0

Tranpose,

2Py\frac{2P}{y}

We will have

dpdy=2Py\frac{dp}{dy}=-\frac{2P}{y}

Integrate both sides,

dpdy=2Py\int \frac{dp}{dy}=-\int\frac{2P}{y}

The equation will become a SEPARABLE DIFFERENTIAL EQUATION, multiply both sides with

dyP\frac{dy}{P}\:

We will come to the equation:

dpP=2ydy \frac{dp}{P}=-\frac{2}{y}dy

Integrate both sides,

dpP=2ydy\int \frac{dp}{P}=-\int\frac{2}{y}dy

The answer will be:

ln(P)=ln(y2)+lnC\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC

Apply logarithmic definition and exponent rule

logab=cthen,b=acab+c=abacloga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c

The answer will be:

P=Cy2P=\frac{C}{y^2}

Recall that

P=dydxP=\frac{dy}{dx}

Substitute the original value of P,

dydx=Cy2\frac{dy}{dx}=\frac{C}{y^2}

Again, this is a Separable Differential Equation, multiply both sides with:

y2dxy^{2}dx

It will become

y2dy=Cdxy^{2}dy=Cdx

Integrate both sides,

y2dy=Cdx\int y^{2}dy=\int Cdx

The answer will be

y33=C1x+C2\frac{y^3}{3}=C1x+C2

Multiply both sides with 3 and the final answer will be

y3=C1x+C2y^3=C_1x+C_2

You can still solve it explicitly,

y=C1x+C23y=\sqrt[3]{C_1x+C_2}

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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 6 — Special Second-Ordered Differential Equations


Solve the following differential equation

yy(y)2+y=0yy''-\left(y'\right)^2+y'=0

Solutions:

Basically, We need to make the orders of each term to 1. To be able to further break down the equation.

soweletP=y             Pdpdy=yso\:we\:let\:P=y' \\ \:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''

Substituting to the equation, we get

yPdPdyP2+P=0yP\frac{dP}{dy}-P^2+P=0

Removing the variables y and P from the 1st term we get

dPdxPy+1y=0      ,theequationhasbecomeaFOLDE  dPdyPy=1y           T(y)=1y,     Q(y)=1y    ϕ=e1ydy=y1         Pϕ=ϕQ(y)dy+C1            Py1=y1(y1)dy+C1Py1=y2dy+C1Py=1y+C1    P=1+yC1dydx=1+yC1\frac{dP}{dx}-\frac{P}{y}+\frac{1}{y}=0\:\:\:\:\:\: , the \:equation\:has\:become\:a \:FOLDE\\ \;\\ \frac{dP}{dy}-\frac{P}{y}=-\frac{1}{y}\:\:\:\:\:\:\:\:\:\:\:T\left(y\right)=-\frac{1}{y},\:\:\:\:\:Q\left(y\right)=-\frac{1}{y}\\ \:\\\:\:\:\: \phi =e^{\int \:-\frac{1}{y}dy}\\ =y^{-1} \\\:\:\:\:\:\:\:\:\:P\phi=\int\phi\:Q(y)dy\:+\:C_{1} \\\:\:\:\:\:\:\:\:\:\:\:\:Py^{-1}=\int \:y^{-1}\left(-y^{-1}\right)dy+C_{1} \\ Py^{-1}=\int \:-y^{-2}dy+C_{1} \\ \:\\ \frac{P}{y}=\frac{1}{y}+C_{1} \\\: \:\\\:\: P=1+yC_{1} \\ \frac{dy}{dx}=1+yC_{1}

By means of Separation of Variables

dy1+yC1=dxdy1+yC1=dx        letu=1+yC1      du=C1dyduC1=dy   1C1duu=dx           1C1lnu=x+C2\\ \frac{dy}{1+yC_{1}}=\:dx \\ \int \:\frac{dy}{1+yC_{1}}=\int \:dx \\\:\:\:\:\:\:\:\: let\:u=1+yC_{1} \\\:\:\:\:\:\: du=C_{1}dy\\ \frac{du}{C_{1}}=dy \\\:\:\: \frac{1}{C_{1}}\int \:\frac{du}{u}=\int \:dx \\\:\:\:\:\:\:\:\:\:\:\: \frac{1}{C_{1}}ln\:u=x+C_{2}

We get

ln1+yC1=C1x+C2ln\left|1+yC_{1}\right|=C_{1}x+C_{2}

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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 2 — Special Second-Ordered Differential Equations


Question:

ddx(xdydx+(1+x)y)=12  ;dydx=0,y=0,x=1\frac{d}{dx}\left(x\:\frac{dy}{dx}\:+\left(1+x\right)\:y\right)\:=12\:\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1

Solution:

Since the equation is in the form d/dx (dy/dx + y P(x) = Q(x) , we use Case 1

letu=xdydx+(1+x)ydudx=12u=12x+C1\begin{align*} let\:u & = x\frac{dy}{dx}+\left(1+x\right)\:y \\ \int \:\frac{du}{dx}\:&=\int \:12 \\u &=12x+C_1 \end{align*}

Solving for the value of C1 using the initial values.

xdydx+(1+x)y=12x+C1;dydx=0,y=0,x=11(0)+(1+1)0=12(1)+C10+0=12+C112=C1\begin{align*} x\frac{dy}{dx}+\left(1+x\right)y\: & =12x\:+C_1\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1 \\1(0) + (1+1)0 & = 12(1) + C_1 \\0+0 & =12+C_1 \\-12 & =C_1 \end{align*}

Rewriting the equation into the general form of a first-order linear differential equation (FOLDE).

[xdydx+(1+x)y=12x12]1xdydx+(1+xx)y=1212xdydx+(1x+1)y=1212x\left[x\frac{dy}{dx}+\left(1+x\right)y=12x-12\right]\frac{1}{x} \\\frac{dy}{dx}+\left(\frac{1+x}{x}\right)y=12-\frac{12}{x} \\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

Since the equation is now in the form of dy/dx + y P(x) = Q(x), we use FOLDE

dydx+(1x+1)y=1212x\\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

From the general form of a first-order differential equation, we have

P(x)=(1x+1)Q(x)=1212x\\P \left( x \right)= \left(\frac{1}{x}+1\right) \\Q\left( x \right)= 12-\frac{12}{x}

Compute for the integrating factor

ϕ=eP(x)dxϕ=e(1x+1)dxϕ=elnx+xϕ=x(ex)\begin{align*} \phi &= e^ {\int P\left(x \right) dx} \\\phi & =e^{\int \:\left(\frac{1}{x}+1\right)dx} \\\phi & \:=e^{\ln x+x} \\\phi & \:=x\left(e^x\right) \end{align*}

Substituting everything to the solution of a first-order linear differential equation, we have

y(xex)=xex(1212x)dx+C2y(xex)=(12xex12ex)dx+C2yxex=12xex12exdx+C2y(xe^x)=\int xe^x\left(12-\frac{12}{x}\right)dx+C_2 \\y\left(xe^x\right)=\int \left(12xe^x-12e^x\right)dx+C_2 \\yxe^x=\int 12xe^x-\int \:12e^x\:dx+C_2​

Use Integration by Parts to solve for the first integral

12xexdx=12xexdxu=xdu=dxdv=ex v=exThereforeuvvdu=xexexdx=xexexConsequently12xexdx=12(xexex)\begin{align*} \int \:12xe^xdx & = 12 \int xe^x dx\\ u = x & &du=dx \\ dv = & e^x \ & v=e^x \\ \text{Therefore} \\ uv-\int \:vdu & =xe^x-\int \:e^xdx \\ & =xe^x-e^x \\ \text{Consequently} \\ \int \:12xe^xdx & = 12 \left( xe^x-e^x\right) \end{align*}

Therefore,

yxex=12(xexex)12ex+C2yxex=12xex12ex12ex+C2yxex=12xex24ex+C2\begin{align*} yxe^x & =12\left(xe^x-e^x\right)-12e^x+C_2 \\yxe^x &=12xe^x-12e^x-12e^x+C_2 \\yxe^x &=12xe^x-24e^x+C_2 \end{align*}

Solving for C2

yx=12x24+C2ex  ;y=0,x=10(1)=12(1)24+C2e10=1224e+C2e10=12+C2e112e1=C2\begin{align*} yx & =12x-24+\frac{C_2}{e^x}\:\:;\:y=0,\:x=1 \\0\left(1\right) & =12\left(1\right)-24+\frac{C_2}{e^1}\: \\0 & =12-24e+\frac{C_2}{e^1}\: \\0 &=-12+\frac{C_2}{e^1}\: \\12e^1 & =C_2\: \end{align*}

Therefore, the solution to the problem is

yx=12x24+12e1exoryx=12x24+12e1xyx=12x-24+\frac{12e^1}{e^x} \\or \\yx=12x-24+12e^{1-x}

Determining if a given Differential Equation is Separable or Not


Determine whether each of the following differential equations is or is not separable, and, if it is separable, rewrite the equation in the form dy/dx=f(x) g(y).
a)dydx=xy3x2y+6\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6
b)dydx=sin(x+y)\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)
c)ydydx=ex3y2\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}


Solution:

Part A

dydx=xy3x2y+6dydx=(xy3x)(2y6)dydx=x(y3)2(y3)dydx=(x2)(y3)\begin{align*} \frac{dy}{dx} & = xy-3x-2y+6 \\ \frac{dy}{dx} & = \left( xy-3x \right)-\left( 2y-6 \right)\\ \frac{dy}{dx} & = x\left( y-3 \right)-2\left( y-3 \right)\\ \frac{dy}{dx} & = \left( x-2 \right)\left( y-3 \right) \end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

Part B

dydx=sin(x+y)dydx=sin(x)cos(y)+cos(x)sin(y)\begin{align*} \frac{dy}{dx} & = \sin\left( x+y \right) \\ \frac{dy}{dx} & = \sin\left( x \right)\cos\left( y \right) +\cos\left( x \right)\sin\left( y \right)\\ \end{align*}

Since F(x,y) is not factorable in the form f(x) g(y), the given differential equation is not separable.

Part C

ydydx=ex3y2ydydx=exe3y2dydx=exye3y2dydx=ex(1ye3y2)\begin{align*} y \frac{dy}{dx} & = e^{x-3y^2}\\ y \frac{dy}{dx} & = \frac{e^x}{e^{3y^2}}\\ \frac{dy}{dx} & =\frac{e^x}{y e^{3y^2}} \\ \frac{dy}{dx} & = e^x \left( \frac{1}{ye^{3y^2}} \right) \\ \end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.