Toronto Math Forum
MAT244-2013F => MAT244 Math--Tests => Quiz 1 => Topic started by: Victor Ivrii on October 03, 2013, 03:18:49 AM
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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 $a(x)z'+b(x)z=0$.
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Here is my solution to this question.
(I used Word to type the math formula. )
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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).
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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
$\Rightarrow 1 = \frac{2u^2}{5t^5} - 2ku^2$
$\Rightarrow 1 = \frac{2}{5t}y^2 - 2ky^2t^4$
$\Rightarrow 1 = y^2(\frac{2}{5t} - 2kt^4)$
$\Rightarrow 1 = y^2(\frac{2 - 10kt^5}{5t})$
$\Rightarrow y^2 = \frac{5t}{2 - 10kt^5}$
$\Rightarrow y = \pm\sqrt{\frac{5t}{2 - 10kt^5}}$
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Obviously (up to the choice of constant) both got the same solution