MAT244-2018S > Quiz-2

Q2-T0601

(1/1)

**Victor Ivrii**:

Using integrating factor method, solve the given initial value problem and determine at least approximately where the solution is valid.

$$

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

$$

**David Chan**:

Let $$M(x, y) = 9x^2 + y - 1 \qquad \text{ and } \qquad N(x, y) = -4y + x$$

Then, $$\frac{\partial}{\partial y}M(x, y) = 1 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 1$$

(Also note that $M$, $N$, $M_y$, $N_x$ are all continuous) Since $M_y = N_x$, the equation is exact. Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 9x^2 + y - 1 = M \tag{1} \\\psi_y(x, y) &= -4y + x = N \tag{2}\end{align*}

Integrating (1) with respect to $x$, we get $$\psi(x, y) = 3x^3 + xy - x + h(y)$$ for some function $h$ of $y$. Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x + h'(y) = -4y + x$$

Therefore, $$h'(y) = -4y \implies h(y) -2y^2$$

and we have $$\psi(x, y) = 3x^3 + xy - x - 2y^2$$

By our choice of $\psi$, we know that $$\frac{\partial \psi}{\partial x} = M \qquad \text{ and } \qquad \frac{\partial \psi}{\partial y} = N$$ so we can rewrite our original differential equation as $$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(\psi(x, y)) = 0$$ Therefore, the (implicit) general solution to out differential equation is $$\psi(x, y) = -2y^2 + xy + 3x^3 - x = C$$

We can solve this explicitly for $y$ to using the quadratic formula to get $$y = \frac{x \pm \sqrt{x^2 + 8(3x^3 - x - C)}}{4}$$

Given our initial condition, $y(1) = 0$, we have $$3(1)^3 - 1 = C \implies C = 2$$

Thus, we have our solution $$y = \frac{x - \sqrt{24x^3 + x^2 - 8x - 16}}{4}, \qquad \text{ valid for } \ 24x^3 + x^2 - 8x - 16 > 0 \iff x > .985$$

Navigation

[0] Message Index

Go to full version