### Author Topic: TT1--Problem 1  (Read 11758 times)

#### Victor Ivrii

• Elder Member
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##### TT1--Problem 1
« on: February 13, 2013, 10:36:58 PM »
Find integrating factor and solve
\begin{equation*}
x\,dx +y (1+x^2+y^2)\,dy=0.
\end{equation*}

#### Matthew Cristoferi-Paolucci

• Jr. Member
• Posts: 10
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##### Re: TT1--Problem 1
« Reply #1 on: February 13, 2013, 10:44:30 PM »
heres my solution

#### Marcia Bianchi

• Newbie
• Posts: 4
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##### Re: TT1--Problem 1
« Reply #2 on: February 13, 2013, 10:46:08 PM »
solution

#### Marcia Bianchi

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##### Re: TT1--Problem 1
« Reply #3 on: February 13, 2013, 10:49:07 PM »

#### Jeong Yeon Yook

• Full Member
• Posts: 20
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##### Re: TT1--Problem 1
« Reply #4 on: February 13, 2013, 10:57:16 PM »
better formatted solution

#### Alexander Jankowski

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• Posts: 23
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##### Re: TT1--Problem 1
« Reply #5 on: February 13, 2013, 11:08:34 PM »
I am late, but at some point, someone is going to want a coded solution that isn't rotated by 90Â° CCW.

Let $M(x,y) = x$ and let $N(x,y) = y + x^2y + y^3$. Then,

\begin{align*}
M_y(x,y) = 0, & N_x(x,y) = 2xy.
\end{align*}

So the equation is not exact, and we need an integrating factor $\mu(x,y) = \mu(y)$ to make it exact.

\begin{equation*}
\frac{d\mu(y)}{dy} = \frac{N_x - M_y}{M} \mu(y) = \frac{2xy}{x} \mu(y) = y\mu(y) \Longrightarrow \frac{d\mu(y)}{\mu(y)} = 2ydy \Longrightarrow \mu(y) = e^{y^2}
\end{equation*}

Now, we require a function $\psi(x,y)$ that satisfies $\psi_x = \mu M$ and $\psi_y = \mu N$.

\begin{equation*}
\psi_x(x,y) = xe^{y^2} \Longrightarrow \psi(x,y) = \frac{1}{2}x^2e^{y^2} + h(y)
\end{equation*}

We differentiate the result to get

\begin{equation*}
\psi_y(x,y) = x^2xe^{y^2} + h'(y) = e^{y^2}(y + x^2y + y^3) \Longleftrightarrow h'(y) = e^{y^2}(y^3 + y) \Longrightarrow h(y) = \int{(e^{y^2}(y^3 + y))}dy.
\end{equation*}

Let $u = y^2$ so that $du = 2ydy$. Then,

\begin{equation*}
h(y) = \int{(y^3e^{y^2})}dy + \int{(ye^{y^2})}dy = \frac{1}{2}\int{(ue^u)}du + \frac{1}{2}\int{e^u}du = \frac{1}{2}ue^u - \frac{1}{2}\int{e^u}du + \frac{1}{2}e^u = \frac{1}{2}ue^u - \frac{1}{2}e^u + \frac{1}{2}e^u = \frac{1}{2}y^2e^{y^2}.
\end{equation*}

Therefore, the solution is implicitly given by

\begin{equation*}
C = \frac{1}{2}e^{y^2}(x^2 + y^2).
\end{equation*}
« Last Edit: February 14, 2013, 12:02:27 AM by Alexander Jankowski »