Differential Equation Exercise 13

We aim to solve the following first-order differential equation:

$$ y' + y = x $$

This is a linear differential equation of the standard form y' + a(x)y = b(x), where $ a(x) = 1 $ and $ b(x) = x $.

To solve it, we’ll apply the Lagrange method, which provides the general solution:

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

Substituting $ a(x) = 1 $ and $ b(x) = x $ into the formula gives:

$$ y = e^{-\int 1 \, dx} \cdot \left[ \int x \cdot e^{\int 1 \, dx} \, dx + c \right] $$

Since $ \int 1 \, dx = x $, we obtain:

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

To evaluate $ \int x \cdot e^x \, dx $, we apply integration by parts, using the identity ∫f g' = f g − ∫f' g, with $ f = x $ and $ g' = e^x $:

$$ \int x \cdot e^x \, dx = x \cdot e^x - \int 1 \cdot e^x \, dx = x \cdot e^x - \int e^x \, dx $$

Since $ \int e^x \, dx = e^x $, we now have:

$$ y = e^{-x} \cdot \left[ x \cdot e^x - e^x + c \right] $$

Distributing the exponential factor $ e^{-x} $ across the expression yields:

$$ y = e^{-x} \cdot x \cdot e^x - e^{-x} \cdot e^x + c \cdot e^{-x} $$

Recognizing that $ e^{-x} \cdot e^x = 1 $, we simplify the expression to:

$$ y = x - 1 + c \cdot e^{-x} $$

This is the general solution of the differential equation:

$$ y = x + c \cdot e^{-x} - 1 $$

And the process can be extended similarly for other cases.