Differential Equation Exercise 18

We are tasked with solving the following first - order differential equation:

$$ 2xyy' = x^2 + 2y^2 $$

To begin, we isolate $y′$ by dividing both sides by $2xy$:

$$ \frac{2xyy'}{2xy} = \frac{x^2 + 2y^2}{2xy} $$

$$ y' = \frac{x^2}{2xy} + \frac{2y^2}{2xy} $$

$$ y' = \frac{x}{2y} + \frac{y}{x} $$

This is a homogeneous differential equation of the form $y′ = f(y/x)$.

To solve it, we introduce the substitution t = y/x:

$$ t = \frac{y}{x} $$

From this, $y$ can be written as:

$$ y = t \cdot x $$

Differentiating with respect to $x$ yields:

$$ D[y] = D[tx] $$

$$ y' = t'x + t $$

Substituting y′ = t′x + t into the original equation gives:

$$ y' = \frac{x}{2y} + \frac{y}{x} $$

$$ t'x + t = \frac{x}{2y} + \frac{y}{x} $$

Replacing $y$ with $tx$:

$$ t'x + t = \frac{1}{2t} + t $$

Writing $t'$ as $\frac{dt}{dx}$:

$$ \frac{dt}{dx} \cdot x + t = \frac{1}{2t} + t $$

Canceling $+t$ from both sides leaves:

$$ \frac{dt}{dx} \cdot x = \frac{1}{2t} $$

At this point, the equation is ready to be solved as a separable differential equation.

Separating variables:

$$ t \, dt = \frac{1}{2x} \, dx $$

Integrating both sides:

$$ \int t \, dt = \int \frac{1}{2x} \, dx $$

$$ \int t \, dt = \frac{1}{2} \int \frac{1}{x} \, dx $$

The integral on the right is $\log(x) + c$:

$$ \int t \, dt = \frac{\log(x)}{2} + c $$

The left-hand side integrates to $\frac{t^2}{2}$:

$$ \frac{t^2}{2} = \frac{\log(x)}{2} + c $$

Multiplying through by 2:

$$ t^2 = \log(x) + c $$

Note: The constant $c$ is left as is. Writing $2c$ or $c$ is equivalent, since $c$ can take any real value, positive or negative.

Substituting back $t = y/x$:

$$ \left(\frac{y}{x}\right)^2 = \log(x) + c $$

$$ \frac{y^2}{x^2} = \log(x) + c $$

Solving for $y$:

$$ y^2 = x^2 \big(\log(x) + c\big) $$

$$ y^2 = x^2 \log(x) + c x^2 $$

Taking square roots:

$$ \sqrt{y^2} = \sqrt{x^2 \log(x) + c x^2} $$

$$ y = \sqrt{x^2 \log(x) + c x^2} $$

This is the general solution of the differential equation.

That completes the solution.