Toronto Math Forum
MAT2442013F => MAT244 MathTests => Quiz 1 => Topic started by: Victor Ivrii on October 03, 2013, 03:19:49 AM

2.6 p 101, # 14
Solve the initial value problem
\begin{equation*}
(9x^2+y1)(4yx)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.

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.