### Author Topic: Q1, P2 Night sections  (Read 2213 times)

#### Victor Ivrii

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##### Q1, P2 Night sections
« on: October 03, 2013, 03:19:49 AM »
2.6 p 101, # 14
Solve the initial value problem
\begin{equation*}
\end{equation*}

Hint: Write an equation in the form $M(x,y)\,dx+N(x,y)\,dy=0$ and check if it is exact.

#### Xuewen Yang

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##### Re: Q1, P2 Night sections
« Reply #1 on: October 03, 2013, 12:52:26 PM »
First I apologize that the format is bad, because it's my first time typing math equations using this forum, so I am not sure how it works.

Typing is always better than scanning. Please try to modify post to see how I did it. Observe special meaning of \$...\$ for inline math and \$\$...\$\$ for display math (occupying its own lines) in the source. -- V.I.

Second, wherever it says Q(x,y), it is not Q actually, it's that Greek letter phi which I don't know how to type. OK, I will change $Q$ to $\Phi$ but I am not sure why you don't like $Q$.
So here we go.

Solution: Rewrite function equation as
$$( 9x^2 + y - 1 ) + ( x - 4y )y' = 0$$
So we have  $M_y(x,y) = 1 = N_x(x,y)$, so this function equation  is exact.
\left\{\begin{aligned} &\Phi_x(x,y) = 9x^2 + y - 1,\\ &\Phi_y(x,y) = x - 4y. \end{aligned}\right.
Above system was more complicated to type.
By integrating $\Phi_x(x,y)$, we have  $\Phi (x,y) = 3x^3 + xy - x + h(y)$ with arbitrary $h(y)$.
By differentiating derived $\Phi (x,y)$ with respect to $y$ we have $\Phi_y(x,y) = x + h'(y)$
Note that   $\Phi_y(x,y) = x + h'(y) = x - 4y$.
Therefore we have  $h'(y) = -4y$, and $h(y) = -2y^2$.
So
$$\Phi (x,y) = 3x^3 + xy - x - 2y^2 = c.$$
Now we plug $y(1) = 0$ into this equation, we have  $3(1^3) + (1)(0) - 1 - 2(0^2) = c\implies c = 2$.
So the solution is
$$3x^3 + xy - x - 2y^2 = 2.$$
Good Job -- V.I.
« Last Edit: October 03, 2013, 05:02:24 PM by Victor Ivrii »