MAT244-2018S > Quiz-2

Q2-T0801

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Victor Ivrii:
Find an integrating factor and solve the given equation.
$$(3x^2y + 2xy + y^3) + (x^2 + y^2)y' = 0.$$

Junya Zhang:
Since the given DE is not exact, we need to find a function $\mu (x,y)$ such that the equation $\mu (3x^{2}y+2xy+y^{3}) + \mu (x^{2}+y^{2})y’=0$ is exact.

This means that
$$\frac{∂}{∂y}[\mu (3x^{2}y+2xy+y^{3})]=\frac{∂}{∂x}[\mu (x^{2}+y^{2})]$$
$$\frac{∂\mu}{∂y}(3x^{2}y+2xy+y^{3})+\mu (3x^{2}+2x+3y^{2}) = \frac{∂\mu}{∂x}(x^{2}+y^{2}) + \mu (2x)$$
$$\frac{∂\mu}{∂y}(3x^{2}y+2xy+y^{3})+\mu (3x^{2}+3y^{2}) = \frac{∂\mu}{∂x}(x^{2}+y^{2})$$

Suppose $\mu$ is a function of $x$ only. (i.e. $\mu=\mu (x)$)
Then $\frac{∂\mu}{∂y}=0$ and $\frac{∂\mu}{∂x}=\frac{d\mu}{dx}$

Then $$\mu (3x^{2}+3y^{2}) = \frac{d\mu}{dx}(x^{2}+y^{2})$$

Divide both sides by $(x^{2}+y^{2})$ we have $$3\mu = \frac{d\mu}{dx}$$ which is a separable equation.

$$\int{\frac{1}{\mu}d\mu} = \int{3 dx}$$
$$ln(\mu)=3x$$
$$\mu = e^{3x}$$

Thus, $\mu = e^{3x}$ is an integration factor for the given DE.

Multiply the given DE by $\mu = e^{3x}$, we get
$$e^{3x} (3x^{2}y+2xy+y^{3}) + e^{3x} (x^{2}+y^{2})y’=0$$ which is exact.

Then by theorem, we know there exist a function $\phi(x,y)=C$ which satisfies the given DE

We also know that
$\frac{∂\phi}{∂x} = e^{3x} (3x^{2}y+2xy+y^{3})$ and $\frac{∂\phi}{∂y} =e^{3x} (x^{2}+y^{2})$

Integrate $\frac{∂\phi}{∂y} =e^{3x} (x^{2}+y^{2})$ with respect to $y$ we have
$$\phi(x,y)= e^{3x} (x^{2}y+\frac{1}{3}y^{3}) +g(x)$$
Then, take derivative on both sides with respect to $x$, we have
$$\frac{∂\phi}{∂x} = 3e^{3x} (x^{2}y+\frac{1}{3}y^{3}) + e^{3x} (2xy) + g’(x)$$
Simplify a little bit we have
$$\frac{∂\phi}{∂x} = e^{3x} (3x^{2}y + y^{3} + 2xy) + g’(x)$$

Since we know that
$\frac{∂\phi}{∂x} = e^{3x} (3x^{2}y+2xy+y^{3})$

This means that $g’(x)=0$. So $g(x)=C$

Altogether, we have $$\phi(x,y)= e^{3x} (x^{2}y+\frac{1}{3}y^{3})=C$$
Which means the general solution to the given DE is $$e^{3x} (x^{2}y+\frac{1}{3}y^{3})=C$$