Differential Equation - Exercise 14

Consider the first-order differential equation:

$$ y' + x y + x = 0 $$

We can rewrite it in the standard linear form $ y' + a(x) y = b(x) $ with $ a(x) = x $ and $ b(x) = -x $:

$$ y' + x y = -x $$

Since this is a first-order linear differential equation, we can solve it using the method of variation of constants (also known as the integrating factor method).

The general solution formula is:

$$ y = e^{-\int a(x) \, dx} \left[ \int b(x) \, e^{\int a(x) \, dx} \, dx + C \right] $$

Substituting $ a(x) = x $ and $ b(x) = -x $:

$$ y = e^{-\int x \, dx} \left[ \int -x \, e^{\int x \, dx} \, dx + C \right] $$

Since $ \int x \, dx = \frac{x^2}{2} $, this becomes:

$$ y = e^{-\frac{x^2}{2}} \left[ -\int x \, e^{\frac{x^2}{2}} \, dx + C \right] $$

The integral $ \int x \, e^{\frac{x^2}{2}} dx $ is straightforward: its antiderivative is $ e^{\frac{x^2}{2}} $, because

Check: $ \frac{d}{dx} \left[ e^{\frac{x^2}{2}} \right] = e^{\frac{x^2}{2}} \cdot \frac{d}{dx} \left[ \frac{x^2}{2} \right] = e^{\frac{x^2}{2}} \cdot x $

Substituting this result gives:

$$ y = e^{-\frac{x^2}{2}} \left[ - e^{\frac{x^2}{2}} + C \right] $$

Multiplying through by $ e^{-\frac{x^2}{2}} $:

$$ y = - e^{-\frac{x^2}{2}} \cdot e^{\frac{x^2}{2}} + C \, e^{-\frac{x^2}{2}} $$

$$ y = -1 + C \, e^{-\frac{x^2}{2}} $$

Thus, the general solution is:

$$ y = C \, e^{-\frac{x^2}{2}} - 1 $$

The two solution methods - direct integration and the integrating factor approach - lead to the same result, confirming its correctness.