Tag Archives: second order differential equations

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=dydx∫ dudx = ∫ (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′′+2y′−6x−3=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=e∫ P(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=6⋅∫ xe2xdx\begin{align*} \int \:e^{2x}6xdx= 6\cdot \int \:xe^{2x}dx \\ \end{align*}

Let u = 2x and du/2 = dx

32∫ euudu\frac{3}{2}\int \:e^uudu

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

nv−∫ndvueu−32∫  eudu=euu−eu=3e2xx−32e2x\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

32∫ eudu=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=3e2xx−32e2x+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+C1e2x∫ dy=∫ (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=3x22−C1e−2x2+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⁡(y−2)+lnC\ln \left(P\right)=\ln \left(y^{-2}\right)+lnC

Apply logarithmic definition and exponent rule

logab=c then, 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.

so we let P=y′             Pdpdy=y′′so\:we\:let\:P=y' \\ \:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''

Substituting to the equation, we get

yPdPdy−P2+P=0yP\frac{dP}{dy}-P^2+P=0

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

dPdx−Py+1y=0      ,the equation has become a FOLDE  dPdy−Py=−1y           T(y)=−1y,     Q(y)=−1y     ϕ=e∫ −1ydy=y−1         Pϕ=∫ϕ Q(y)dy + C1            Py−1=∫ y−1(−y−1)dy+C1Py−1=∫ −y−2dy+C1 Py=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= dx∫ dy1+yC1=∫ dx        let u=1+yC1      du=C1dyduC1=dy   1C1∫ duu=∫ dx           1C1ln u=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

ln∣1+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(x dydx +(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

let u=xdydx+(1+x) y∫ dudx =∫ 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+C1−12=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=12x−12]1xdydx+(1+xx)y=12−12xdydx+(1x+1)y=12−12x\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=12−12x\\\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)=12−12x\\P \left( x \right)= \left(\frac{1}{x}+1\right) \\Q\left( x \right)= 12-\frac{12}{x}

Compute for the integrating factor

ϕ=e∫P(x)dxϕ=e∫ (1x+1)dxϕ =eln⁡x+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(12−12x)dx+C2y(xex)=∫(12xex−12ex)dx+C2yxex=∫12xex−∫ 12ex dx+C2​y(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=12∫xexdxu=xdu=dxdv=ex v=exThereforeuv−∫ vdu=xex−∫ exdx=xex−exConsequently∫ 12xexdx=12(xex−ex)\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(xex−ex)−12ex+C2yxex=12xex−12ex−12ex+C2yxex=12xex−24ex+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=12x−24+C2ex  ; y=0, x=10(1)=12(1)−24+C2e1 0=12−24e+C2e1 0=−12+C2e1 12e1=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=12x−24+12e1exoryx=12x−24+12e1−xyx=12x-24+\frac{12e^1}{e^x} \\or \\yx=12x-24+12e^{1-x}

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

Find the general solution of the differential equation

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

Solution:

ddx(dydx)=6x+3 let u=dydx dudx=6x+3 Integrate,∫dudx=∫(6x+3)dx ∫dudx=6∫xdx+3∫dx u=6x22+3x+C1 u=3x2+3x+C1 Substitute,dydx=3x2+3x+C1 dy=(3x2+3x+C1)dx Integrate,∫dy=∫(3x2+3x+C1)dx ∫dy=3∫x2dx+3∫xdx+C1∫dx y=3x33+3x22+C1x+C2 Simplify,y=x3+3x22+C1x+C2\begin{align*} \frac{d}{dx}\left(\frac{dy}{dx}\right) & =6x+3 \\\ \\ let\:u & =\frac{dy}{dx} \\\ \\ \frac{du}{dx} & =6x+3 \\\ \\ Integrate,\\ \int \frac{du}{dx} & =\int (6x+3)dx \\\ \\ \int \frac{du}{dx} & =6\int xdx+3\int dx \\\ \\ u & =\frac{6x^2}{2}+3x+C_1 \\\ \\ u & =3x^2+3x+C_1 \\\ \\ Substitute, \\ \frac{dy}{dx} & =3x^2+3x+C_1 \\\ \\ dy & =\left(3x^2+3x+C_1\right)dx \\\ \\ Integrate,\\ \int dy & =\int (3x^2+3x+C_1)dx \\\ \\ \int dy & =3\int x^2dx+3\int xdx+C_1\int dx \\\ \\ y & =\frac{3x^3}{3}+\frac{3x^2}{2}+C_1x+C_2 \\\ \\ Simplify, \\ y & =x^3+\frac{3x^2}{2}+C_1x+C_2 \\ \end{align*}
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