First-Order Differential Equation by Substitution

Question

Solve the differential equation by the appropriate substitution $$x^2\frac{dy}{dx}+2xy=x^4y^2+1$$ .

Answer 1

Use the product rule for derivatives to find the pattern you need.

$$x^2 y' + 2 x y = \frac{d}{dx}[\ldots]$$

Big hint is that

$$x^4 y^2 = (x^2 y)^2$$

ADDENDUM

Now that the cat's out of the bag, I may as well show the OP that this equation is a nice, simple, linear equation that does not require techniques from Riccati equations.

My hint was for the OP to observe that

$$x^2 y' + 2 x y = \frac{d}{dx} [x^2 y]$$

so that the equation may be rewritten as

$$\frac{d}{dx} [x^2 y] = 1+ (x^2 y)^2$$

or

$$\frac{d(x^2 y)}{1 + (x^2 y)^2} = dx$$

Integrate both sides; recalling that

$$\int \frac{du}{1+u^2} = \arctan{u}$$

we have

$$\arctan{(x^2 y)} = x + C$$

where $C$ is a constant of integration. Taking the tangent of both sides and dividing by $x^2$, we get

$$y = \frac{\tan{(x + C)}}{x^2}$$

Answer 2

$x^2\dfrac{dy}{dx}+2xy=x^4y^2+1$

$x^2\dfrac{dy}{dx}=x^4y^2-2xy+1$

$\dfrac{dy}{dx}=x^2y^2-\dfrac{2y}{x}+\dfrac{1}{x^2}$

Note that this belongs to a Riccati equation and cannot find an obvious particular solution, so we should follow the method in http://en.wikipedia.org/wiki/Riccati_equation#Reduction_to_a_second_order_linear_equation.

Let $u=x^2y$ ,

Then $y=\dfrac{u}{x^2}$

$\dfrac{dy}{dx}=\dfrac{1}{x^2}\dfrac{du}{dx}-\dfrac{2u}{x^3}$

$\therefore\dfrac{1}{x^2}\dfrac{du}{dx}-\dfrac{2u}{x^3}=x^2\left(\dfrac{u}{x^2}\right)^2-\dfrac{2}{x}\dfrac{u}{x^2}+\dfrac{1}{x^2}$

$\dfrac{1}{x^2}\dfrac{du}{dx}-\dfrac{2u}{x^3}=\dfrac{u^2}{x^2}-\dfrac{2u}{x^3}+\dfrac{1}{x^2}$

$\dfrac{1}{x^2}\dfrac{du}{dx}=\dfrac{u^2}{x^2}+\dfrac{1}{x^2}$

$\dfrac{du}{dx}=u^2+1$

But luckily we get a separable ODE.

$\dfrac{du}{u^2+1}=dx$

$\int\dfrac{du}{u^2+1}=\int dx$

$\tan^{-1}u=x+C$

$u=\tan(x+C)$

$\therefore y=\dfrac{\tan(x+C)}{x^2}$