Differential Equation Exercise 1

We want to solve the following differential equation:

$$ 2x (y^2+1) - y' = 0 $$

This is a first-order differential equation because the highest derivative that appears is the first derivative $ y' $.

It is not linear, since the unknown function appears squared as $ y^2 $.

Our first step is to rewrite the equation in standard form, isolating $ y' $ in terms of the other variables:

$$ y' = 2x (y^2+1) $$

From this form, it is clear that the equation has the structure $ y' = f(x) \cdot g(y) $, which can be solved using the method of separation of variables, where $ f(x) = 2x $ and $ g(y) = y^2 + 1 $.

Using Leibniz notation, $ y' = \frac{dy}{dx} $:

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

We now separate the variables, placing all terms in $ y $ on one side and all terms in $ x $ on the other:

$$ \frac{dy}{y^2+1} = 2x \, dx $$

Next, integrate both sides with respect to their respective variables:

$$ \int \frac{dy}{y^2+1} = \int 2x \, dx $$

$$ \int \frac{1}{y^2+1} \, dy = \int 2x \, dx $$

The antiderivative on the left is $ \arctan(y) + c $:

$$ \arctan(y) + c_1 = \int 2x \, dx $$

Note. The derivative of the arctangent is $$ D[ \arctan(y) ] = \frac{1}{1+y^2} $$

The antiderivative on the right is $ x^2 + c $:

$$ \arctan(y) + c_1 = x^2 + c_2 $$

Combining constants into a single constant $ c = c_1 + c_2 $, we have:

$$ \arctan y = x^2 + c $$

Applying the tangent function to both sides gives:

$$ \tan(\arctan y) = \tan(x^2 + c) $$

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

Note. The tangent function is the inverse of the arctangent, so $ \tan(\arctan y) = y $.

Therefore, the general solution of the differential equation is:

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

That completes the solution.