Author Archives: nikidomingo

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(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.

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

\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) 

\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,

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

By integrating,

\begin{align*}

\int \:ydx&=\int \:xdy\\
\frac{y^2}{2}&=\frac{x^2}{2}+C\\
y^2&={x^2}+2C\\
\end{align*}

We get,

y^2-x^2-=2C
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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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