MAT244--2019F > Quiz-2

TUT0602 QUIZ 2

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**Fenglun Wu**:

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.

$$x^2y^3 + x(1+y^2)y' = 0, ~~~~ \mu(x, y) = \frac{1}{xy^3}$$

First, let's show the given equation isn not exact.

Define $M(x, y) = x^2y^3, ~~ N(x, y) = x(1+y^2)$

$$M_y = \frac{\partial}{\partial y}[x^2y^3] = 3x^2y^2$$

$$N_x = \frac{\partial}{\partial x}[x(1+y^2)] = 1 + y^2$$

Since $M_y \neq N_x$, this implies that the given equation is not exact.

Next, show that the given equation multiplied by the integrating factor $\mu(x, y) = \frac{1}{xy^3}$ is exact.

The new equation becomes

$$ \frac{1}{xy^3}x^2y^3 + \frac{1}{xy^3}x(1+y^2)y' = x + (y^{-3} + y^{-1})y' =0 $$

Define $M'(x, y) = x, ~~ N'(x, y) = y^{-3} + y^{-1}$

$$M'_y = \frac{\partial}{\partial y}(x) = 0$$

$$N'_x = \frac{\partial}{\partial x}(y^{-3} + y^{-1}) = 0$$

Since $M'_y = N'_x$, this implies that the given equation is exact.

Thus, we know that there exists a function $\phi (x, y) = C$ which satisfies the give differential equation.

Also,

$$\frac{\partial \phi}{\partial x} = M'(x, y) = x$$

$$\frac{\partial \phi}{\partial y} = N'(x, y) = y^{-3} + y^{-1}$$

Integrate $\frac{\partial \phi}{\partial x} = x$ with respect to $x$ we have

$$\phi (x, y) = \frac{1}{2}x^2 + h(y)$$

Take derivative on both sides with respect to $y$ we get

$$\frac{\partial \phi}{\partial y} = h'(y)$$

Since we know $\frac{\partial \phi}{\partial y} = N'(x, y) = y^{-3} + y^{-1}$

Then $$ h'(y) = y^{-3} + y^{-1}$$

Integrate $h'(y)$ with respect to y we have

$$ h(y) = -\frac{1}{2}y^{-2} + ln|y| + C$$

Therefore, we have

$$ \phi (x, y) = \frac{1}{2}x^{2} - \frac{1}{2}y^{-2} + ln|y| = C$$

is the general solution to the given differential equation.

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