### Author Topic: Q1: TUT 0101  (Read 2515 times)

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### 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

• Elder Member
• Posts: 2599
• Karma: 0
##### 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

• Jr. Member
• Posts: 14
• Karma: 0
##### 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

• Jr. Member
• Posts: 6
• Karma: 8
##### 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 »