MAT244-2013F > Quiz 1

Q1, P2 Night sections

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**Victor Ivrii**:

2.6 p 101, # 14

Solve the initial value problem

\begin{equation*}

(9x^2+y-1)-(4y-x)y'=0,\qquad y(1)=0.

\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**:

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 $Q$. :D

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.

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