Question

1.y(3x^3-x+y)dx+x^2(1-x^2)dy=0
2.y(2x+y^2)dx+x(y^2-x)dy=0

Answer 1

1.

Solution;

Rewrite the solution as;

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

$\frac{dy}{dx}x^2(1-x^2)+y(3x^3-x)=-y^2$

Divide by $x^2(1-x^2)$

$\frac{dy}{dx}+\frac{y(3x^3-x}{x^2(1-x^2}=\frac{-y^2}{x^2(1-x^4}$

Divide by -y²

$-\frac{dy}{dx}+\frac{(1-3x^2)}{(x-x^3)y}=\frac{1}{x^2-x^4}$

Let v=$\frac 1y$ so that $\frac{dv}{dx}=\frac{-dy}{dxy^2}$

Substitute in the equation;

$\frac{dv}{dx}+\frac{(1-3x^2)v}{x-x^3}$ =$\frac{1}{x^2-x^4}$

Find the integrating factor;

I.F=$e^{\int{\frac{1-3x^2}{x-x^3}}}=e^{ln(x^3-x)}=x^3-x$

Multiply all through with the integrating factor;

Substitute 3x²-1=$\frac{d(x^3-x)}{dx}$

$(x^3-x)\frac{dv}{dx}+\frac{d(x^3-x)v}{dx}=\frac{x^3-x}{x^2-x^4}$

Apply reverse product rule:$\frac{f(dg)}{dx}+\frac{g(df)}{dx}=\frac{d(fg)}{dx}$

We have;

$\frac{d(x^3-x)v}{dx}=\frac{x^3-x}{x^2-x^4}$

Integrate both sides with respect to x;

($x^3-x)v=-ln(x)+c$

$v=\frac{-ln(x)+C}{x^3-x}$

But $v=\frac1y$

$\frac1y=\frac{ln(x)+C}{x-x^3}$

Simplify;

y=$\frac{x-x^3}{ln(x)+C}$

2.

Solution;

By distribution,the equation may be written as;

$(2xy+y^3)dx+(xy^2-x^2)dy=0$

Check if the equation is exact,

M=$2xy+y^3 ;\frac{dM}{dy}=2x+3y^2$

N=$xy^2-x^2 ;\frac{dN}{dx}=y^2-2x$

$\frac{dM}{dy}\neq\frac{dN}{dx}$ ,the equation is not exact.

Mx-Ny=2x²y+y³x-xy³+x²y=3x²y

Take an integrating factor given as;

I.F=$\frac{1}{Mx-Ny}=\frac{1}{3x^2y}$

Multiply the given equation with the integrating factor ;

$\frac{2x+y^2}{3x^2}dx+\frac{y^2-x}{3xy}dy=0$

The general solution of the problem will be given by;

$\int{Mdx}+\int$(Terms of N which are independent of x)

$\int{\frac{2x+y^2}{3x^2}}dx+\int{\frac{-1}{3y}}dy=C$

$\frac{2ln(x)}{3}-\frac{y^2}{3x}-\frac{ln(y)}{3}=C$