91Ó°ÊÓ

We have the given first-order non-exact differential equation as: $$ \left(2 x y+y^{2}\right) d x+\left(2 x y+x^{2}-2 x^{2} y^{2}-2 x y^{3}\right) d y=0 $$ **Step 2: Identify the possible integrating factors**

With the given equation, we have: $$ M(x, y) = 2xy + y^2 \quad \text{and} \quad N(x, y) = 2xy+x^2-2 x^{2} y^{2}-2 x y^{3} $$ Consider an integrating factor, µ(x), a function of only x. We want µ(x) such that the new equation is exact, meaning that: $$ \frac{\partial}{\partial y} \left(\mu(x) M(x, y)\right) = \frac{\partial}{\partial x} \left(\mu(x) N(x, y)\right) $$ **Step 3: Calculate the necessary derivatives and integrating factor**

First, calculate the partial derivatives: $$ \frac{\partial}{\partial y} \left(\mu(x) M(x, y)\right) = \mu(x) \frac{\partial M}{\partial y} = \mu(x) (2x+2y) $$ $$ \frac{\partial}{\partial x} \left(\mu(x) N(x, y)\right) = \mu'(x)N(x, y) + \mu(x) \frac{\partial N}{\partial x} = \mu'(x)(2xy+x^2-2 x^{2} y^{2}-2 x y^{3}) + \mu(x)(2y + 2x - 4x^2 y^2 - 6xy^3). $$ Now, equate the two partial derivatives: $$ \mu(x) (2x+2y) = \mu'(x)(2xy+x^2-2 x^{2} y^{2}-2 x y^{3}) + \mu(x)(2y + 2x - 4x^2 y^2 - 6xy^3) $$ Divide both sides by \(\mu(x)(2xy + y^2)\) to separate the variables: $$ \frac{2x+2y}{2xy + y^2} = \frac{\mu'(x)}{\mu(x)} \frac{(2xy+x^2-2 x^{2} y^{2}-2 x y^{3})}{(2y + 2x - 4x^2 y^2 - 6xy^3)} $$ Now we can focus on the x-part since µ(x) is a function of x only. On the x-part: $$ \frac{2x}{2xy + y^2} = \frac{\mu'(x)}{\mu(x)} \frac{(2xy+x^2-2 x^{2} y^{2}-2 x y^{3})}{(2y + 2x - 4x^2 y^2 - 6xy^3)} $$ We can now integrate with respect to x on both sides: $$ \int \frac{2x}{2xy + y^2} dx = \int \frac{\mu'(x)}{\mu(x)} \frac{(2xy+x^2-2 x^{2} y^{2}-2 x y^{3})}{(2y + 2x - 4x^2 y^2 - 6xy^3)} dx $$ This gets us: $$ \int \frac{1}{x} dx = \int \frac{\mu'(x)}{\mu(x)} dx $$ Integrating both sides with respect to x: $$ \ln(x) + C_1 = \ln\left(\mu(x)\right) + C_2 $$ Expanding and isolating \(\mu(x)\): $$ \mu(x) = xe^C = kx $$ where k is an arbitrary constant. **Step 4: Multiply by the integrating factor and solve the exact differential equation**

Multiply the original equation by the integrating factor \(kx\) and get: $$ kx\left(2 x y+y^{2}\right) d x + kx\left(2 x y+x^{2}-2 x^{2} y^{2}-2 x y^{3}\right) d y=0 $$ The equation becomes: $$ (2 k x^2 y+k x^3 y^{2}) d x + (2 k x^2 y+k x^3 x^{2}-2 k x^{4} y^{2}-2 k x^3 y^{3}) dy=0 $$ Now we have an exact equation of the form: $$ P(x, y) dx + Q(x, y) dy = 0 $$ where \(P(x, y) = 2 k x^2 y+k x^3 y^{2}\) and \(Q(x, y) = 2 k x^2 y+k x^3 x^{2}-2 k x^{4} y^{2}-2 k x^3 y^{3}\). Now, we look for a potential function F(x,y) such that: $$ \frac{\partial F}{\partial x} = P(x, y) \quad \text{and} \quad \frac{\partial F}{\partial y} = Q(x, y) $$ Integrate \(P(x, y)\) with respect to x: $$ F(x, y) = \int P(x, y) dx = \int (2 k x^2 y+k x^3 y^{2}) dx = kx^3 y + \frac{k}{2}x^4 y^2 + C(y) $$ Now, differentiate F(x, y) with respect to y: $$ \frac{\partial F}{\partial y} = kx^3 + kx^4 y + \frac{dC(y)}{dy} $$ Equating this to Q(x, y) and solving for \(\frac{dC(y)}{dy}\): $$ \frac{dC(y)}{dy} = 2 k x^2 y + k x^3 x^{2}-2 k x^{4} y^{2}-2 k x^3 y^{3} - (kx^3 + kx^4 y) $$ $$ \frac{dC(y)}{dy} = 2 k x^2 y - k x^4 y - 2 k x^3 y^{3} $$ Integrate with respect to y: $$ C(y) = 2 k x^2 \int y dy - k x^4 \int y dy - 2 k x^3 \int y^{3} dy = ky^2 x^2 - \frac{1}{2}kx^4y^2 - \frac{1}{2}kx^3y^4 + C $$ The potential function F(x,y) is given by: $$ F(x, y) = kx^3 y + \frac{k}{2}x^4 y^2 + ky^2 x^2 - \frac{1}{2}kx^4y^2 - \frac{1}{2}kx^3y^4+C $$ Since the exact equation must have \(F(x, y) = C\), we have: $$ kx^3 y + \frac{k}{2}x^4 y^2 + ky^2 x^2 - \frac{1}{2}kx^4y^2 - \frac{1}{2}kx^3y^4 = C $$ Simplifying, the solution to the differential equation is: $$ x^3 y + y^2 x^2 - \frac{1}{2}x^3y^4 = C $$