### Author Topic: Q1, P1 Night sections  (Read 3330 times)

#### Victor Ivrii

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##### Q1, P1 Night sections
« on: October 03, 2013, 03:18:49 AM »
2.4 p. 78, # 28
Solve Bernoulli equation
\begin{equation*}
t^2 y' +2ty-y^3=0
\end{equation*}

Hint: To solve Bernoulli equation $a(x)y'+b(x)y+c(x)y^n=0$ with $n\ne 0,1$  you may  either reduce it to the linear equation by substitution  $u=y^{1-n}$ or  to equation with separable variables by substitution $y=zu$ where $z$ is a solution of the corresponding linear homogeneous equation $a(x)z'+b(x)z=0$.

#### Huaqing Fu

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##### Re: Q1, P1 Night sections
« Reply #1 on: October 03, 2013, 07:23:23 PM »
Here is my solution to this question.
(I used Word to type the math formula. )

#### Victor Ivrii

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##### Re: Q1, P1 Night sections
« Reply #2 on: October 03, 2013, 07:38:41 PM »

I am not sure how $u$ became $v$ and then we have also $\mathsf{u}$.

Since you obviously typed your equation, probably into MSW why have not you typed it here? What you posted cannot be copied, recycled and reused. BTW MSW also uses (a kind of) LaTeX internally for math snippets (it was not the case several years ago).
« Last Edit: October 04, 2013, 08:40:19 AM by Victor Ivrii »

#### Terry Ta

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##### Re: Q1, P1 Night sections
« Reply #3 on: October 04, 2013, 07:26:42 AM »
I will show my solution using the second method given in the hint.

First, we will find $z$.

$a(t)z' + b(t)z = 0$
$\Rightarrow t^2z' + 2tz = 0$
$\Rightarrow (t^2z)' = 0$
$\Rightarrow t^2z = c$, where $c$ is an arbitrary constant
$\Rightarrow z = \frac{c}{t^2}$

I will choose $c$ to be 1 and substitute $y = zu = \frac{1}{t^2}u$. Then $\frac{dy}{dt} = -\frac{2}{t^3}u + \frac{1}{t^2}\frac{du}{dt}$.
From the Bernoulli equation,

$t^2y' + 2ty - y^3 = 0$
$\Rightarrow t^2(-\frac{2}{t^3}u + \frac{1}{t^2}\frac{du}{dt}) + 2t(\frac{1}{t^2}u) - (\frac{1}{t^2}u)^3 = 0$
$\Rightarrow -\frac{2u}{t} + \frac{du}{dt} + \frac{2u}{t} - \frac{u^3}{t^6} = 0$
$\Rightarrow \frac{du}{dt} = \frac{u^3}{t^6}$
$\Rightarrow \frac{1}{u^3}du = \frac{1}{t^6}dt$

The last equation above is separable. Integrating the left side with respect to $u$ and the right side with respect to $t$ gives us

$-\frac{1}{2u^2} = -\frac{1}{5t^5} + k$, where $k$ is an arbitrary constant
$\Rightarrow 1 = \frac{2u^2}{5t^5} - 2ku^2$
$\Rightarrow 1 = \frac{2}{5t}y^2 - 2ky^2t^4$
$\Rightarrow 1 = y^2(\frac{2}{5t} - 2kt^4)$
$\Rightarrow 1 = y^2(\frac{2 - 10kt^5}{5t})$
$\Rightarrow y^2 = \frac{5t}{2 - 10kt^5}$
$\Rightarrow y = \pm\sqrt{\frac{5t}{2 - 10kt^5}}$

#### Victor Ivrii

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##### Re: Q1, P1 Night sections
« Reply #4 on: October 04, 2013, 08:41:40 AM »
Obviously (up to the choice of constant) both got the same solution