Q19Mathematics II

Question

Q.2 Solve -

(x^{4} y^{4} + x^{2} y^{2} + x y) y d x + (x^{4} y^{4} - x^{2} y^{2} + x y) x d y = 0

Answer 1

The given differential equation strictly takes the standard form $M d x + N d y = 0$ :

M = x^{4} y^{5} + x^{2} y^{3} + x y^{2}

N = x^{5} y^{4} - x^{3} y^{2} + x^{2} y

Step 1: Finding the Integrating Factor (I.F.)

This specific equation strictly matches the specialized mathematical form $y f_{1} (x y) d x + x f_{2} (x y) d y = 0$ . For equations exactly of this type, the fundamental Integrating Factor is rigidly defined as $I . F . = \frac{1}{M x - N y}$ (provided $M x - N y \neq = 0$ ).

We rigorously calculate the denominator:

M x = x (x^{4} y^{5} + x^{2} y^{3} + x y^{2}) = x^{5} y^{5} + x^{3} y^{3} + x^{2} y^{2}

N y = y (x^{5} y^{4} - x^{3} y^{2} + x^{2} y) = x^{5} y^{5} - x^{3} y^{3} + x^{2} y^{2}

M x - N y = (x^{5} y^{5} + x^{3} y^{3} + x^{2} y^{2}) - (x^{5} y^{5} - x^{3} y^{3} + x^{2} y^{2}) = 2 x^{3} y^{3}

I.F. = \frac{1}{2 x ^{3} y ^{3}}

Step 2: Making the Equation Exact

We systematically multiply the entire original ODE exactly by this I.F. to produce a strictly exact equation $M^{'} d x + N^{'} d y = 0$ :

M^{'} = M \times I.F. = \frac{x ^{4} y ^{5} + x ^{2} y ^{3} + x y ^{2}}{2 x ^{3} y ^{3}} = \frac{1}{2} (x y^{2} + \frac{1}{x} + \frac{1}{x ^{2} y})

N^{'} = N \times I.F. = \frac{x ^{5} y ^{4} - x ^{3} y ^{2} + x ^{2} y}{2 x ^{3} y ^{3}} = \frac{1}{2} (x^{2} y - \frac{1}{y} + \frac{1}{x y ^{2}})

Step 3: Integrating the Exact Equation

For a strictly exact differential equation, the absolute general solution is rigorously found entirely by the standard formula:

\int M^{'} d x (treating y as constant) + \int N^{'} d y (only terms strictly without x) = C

Integrating $M^{'}$ with respect to $x$ :

\int \frac{1}{2} (x y^{2} + \frac{1}{x} + x^{- 2} y^{- 1}) d x = \frac{1}{2} (\frac{x ^{2} y ^{2}}{2} + ln x - \frac{x ^{- 1}}{y})

Integrating terms in $N^{'}$ explicitly without $x$ . The only such term is $- \frac{1}{2 y}$ :

\int - \frac{1}{2 y} d y = - \frac{1}{2} ln y

Combining these parts rigorously yields the absolute final solution:

\frac{x ^{2} y ^{2}}{4} + \frac{1}{2} ln x - \frac{1}{2 x y} - \frac{1}{2} ln y = C

\frac{x ^{2} y ^{2}}{4} + \frac{1}{2} ln (\frac{x}{y}) - \frac{1}{2 x y} = C

Q19Mathematics II

Question

Answer

Step 1: Finding the Integrating Factor (I.F.)

Step 2: Making the Equation Exact

Step 3: Integrating the Exact Equation