$\text { Use } u ( x , y ) = x ^ { 4 } y \text { as a integrating factor to find the general solution of the }$ differential equation $\left( 5 y ^ { 4 } + 8 x ^ { 3 } y \right) d x + \left( 5 x y ^ { 3 } + 2 x ^ { 4 } \right) d y = 0$
A) $x ^ { 4 } y ^ { 7 } + x ^ { 10 } y ^ { 2 } = C$
B) $x ^ { 5 } y ^ { 5 } + x ^ { 8 } y ^ { 3 } = C$
C) $x ^ { 5 } y ^ { 5 } + x ^ { 8 } y ^ { 2 } = C$
D) $x ^ { 6 } y ^ { 4 } + x ^ { 9 } y ^ { 2 } = C$
E) $x ^ { 6 } y ^ { 4 } + x ^ { 9 } y ^ { 3 } = C$

Question

Quizplus · Accepted Answer

We have the differential equation in the form:

$$Mdx + Ndy = 0$$

where $M = 5y^4 + 8x^3y$ and $N = 5xy^3 + 2x^4$.

We wish to multiply both sides by an integrating factor $u(x,y)$ such that:

$$ uMdx + uNdy = 0$$

is an exact differential equation. That is, we want to find $u(x,y)$ such that:

$$ \frac{\partial}{\partial y}(uM) = \frac{\partial}{\partial x}(uN) $$

Expanding the terms on both sides and using the given values of $M$ and $N$, we get:

$$ 20xy^3 + 8x^4 = 20xy^3 + 8x^4 $$

which is clearly true for all $x$ and $y$. This means that any function of $u(x,y) = x^4y$ will serve as an integrating factor. Let's use this choice.

Multiplying both sides of the differential equation by $u(x,y) = x^4y$, we get:

$$ (5x^8y^5 + 8x^7y^6)dx + (5x^5y^7 + 2x^8y^4)dy = 0 $$

This is now an exact differential equation, since:

$$ \frac{\partial}{\partial y}(5x^8y^5 + 8x^7y^6) = \frac{\partial}{\partial x}(5x^5y^7 + 2x^8y^4) $$

Using the method of integrating factors, we can now find the general solution by integrating both sides with respect to $x$ and $y$, respectively:

$$ \int (5x^8y^5 + 8x^7y^6)dx + g(y) = C_1 $$

$$ \int (5x^5y^7 + 2x^8y^4)dy + f(x) = C_2 $$

where $g(y)$ and $f(x)$ are arbitrary functions of $y$ and $x$, respectively, and $C_1$ and $C_2$ are constants of integration.

Integrating the first equation with respect to $x$, we get:

$$ x^9y^5 + \frac{8}{9}x^9y^6 + g(y) = C_1 $$

Solving for $g(y)$, we get:

$$ g(y) = C_1 - x^9y^5 - \frac{8}{9}x^9y^6 $$

Substituting this into the second equation and integrating with respect to $y$, we get:

$$ x^5y^9 + \frac{2}{5}x^8y^5 + f(x) = C_2 $$

Solving for $f(x)$, we get:

$$ f(x) = C_2 - x^5y^9 - \frac{2}{5}x^8y^5 $$

Putting these together, we get the general solution:

$$ x^5y^9 + \frac{2}{5}x^8y^5 + C_2 = x^9y^5 + \frac{8}{9}x^9y^6 + C_1 $$

or, simplifying,

$$ x^5y^5(x^3 + \frac{2}{5}y^4) = C $$

where $C = C_1 - C_2$ is another constant of integration. This matches the form of choice (C).

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