Author Topic: Q1: TUT 0101  (Read 1466 times)

Victor Ivrii

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Q1: TUT 0101
« on: September 28, 2018, 03:26:52 PM »
Find the general solution of the given differential equation:
\begin{equation*}
\frac{dy}{dx}= \frac{x^2+3y^2}{2xy}
\end{equation*}

Victor Ivrii

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Re: Q1: TUT 0101
« Reply #1 on: September 29, 2018, 02:43:40 PM »
No need to post basically  the same solution as the person before you. And definitely if you want to put scanned solution:
1) NO no-ASCII names.
2) It should be well-written

Sorry, deleted both

Nikita Dua

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Re: Q1: TUT 0101
« Reply #2 on: September 29, 2018, 06:15:05 PM »
My solution
« Last Edit: September 29, 2018, 06:27:52 PM by Nikita Dua »

Yiting Zhang

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Re: Q1: TUT 0101
« Reply #3 on: September 29, 2018, 06:56:03 PM »
$$
\begin{split}
y' &= \frac{x^2+3y^2}{2xy} \\
&= \frac{x}{2y} + \frac{3y}{2x} \\
&= \frac{1}{2} (\frac{y}{x})^{-1} + \frac{3}{2}(\frac{y}{x})
\end{split}
$$

$$v = \frac{y}{x} \implies y = vx$$

$$
\begin{split}
y' &= v'x+ v \\
&= \frac{1}{2} v^{-1} + \frac{3}{2} v \\
\end{split}
$$

$$
\begin{split}
v'x &= y' - v \\
&= \frac{1}{2} v^{-1} + \frac{1}{2} v \\
&= \frac{1+v^2}{2v}
\end{split}
$$

$$
\begin{split}
\frac{2v}{1+v^2} dv &= \frac{1}{x} dx \\
\ln |1+v^2| &= \ln |x| + c \\
1+v^2 &= Cx, C = e^c \\
\end{split}
$$

$$
\begin{split}
1 + \frac{y^2}{x^2} - Cx &= 0\\
y^2 + x^2 - Cx^3 &= 0
\end{split}
$$


« Last Edit: September 29, 2018, 07:06:40 PM by Yiting Zhang »