Tag Archives: special second ordered differential equations

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

Find the General Solution

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

Solution:

\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=0

Solution:

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

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+3

ii. Let 

\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.

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

Find the integrating factor

\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

\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

\begin{align*}
\int \:e^{2x}6xdx=

6\cdot \int \:xe^{2x}dx
\\
\end{align*}

Let u = 2x and du/2 = dx

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

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

\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

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

Let u = 2x and du/2 = dx

\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

\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,

\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=\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\left(y\right)^2=0

Solution:

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

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

Denote and substitute to the given equation.

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

We will have,

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

Divide both sides with

 \:\frac{1}{yP}

We will come to,

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

Tranpose, 

\frac{2P}{y}

We will have

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

Integrate both sides,

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

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

\frac{dy}{P}\:

We will come to the equation:

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

Integrate both sides,

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

The answer will be:

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

Apply logarithmic definition and exponent rule

loga^b=c\:then,\:b=a^c\\a^{b+c}=a^ba^c

The answer will be:

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

Recall that

P=\frac{dy}{dx}

Substitute the original value of P,

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

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

y^{2}dx

It will become

y^{2}dy=Cdx

Integrate both sides,

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

The answer will be

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

Multiply both sides with 3 and the final answer will be

y^3=C_1x+C_2

You can still solve it explicitly,

y=\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''-\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' \\
\:\:\:\:\:\:\:\:\:\:\:\:\:P\frac{dp}{dy}=y''

Substituting to the equation, we get 

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

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

\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

\\
\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\left|1+yC_{1}\right|=C_{1}x+C_{2}
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