91Ó°ÊÓ

Initial conditions play a crucial role in solving differential equations. They help us determine particular solutions from a general solution. When we have a differential equation, it often has a family of solutions.
An initial condition is an additional piece of information that allows us to find the exact solution unique to that problem.
In our exercise, the initial condition is given as \( y(0) = 1 \). This means that when \( x = 0 \), the value of \( y \) must be 1.
We use this condition to determine the constant \( B \) in the general solution form \( y(x) = A + B e^{x^2/2} \).

• First, substitute \( x = 0 \) into the expression \( y(x) = A + B e^{x^2/2} \).
• Since \( e^{0^2/2} = 1 \), the equation simplifies to \( y(0) = A + B \).
• The initial condition \( y(0) = 1 \) then becomes \( A + B = 1 \).

Using this step, we find the value of \( B \) after determining \( A \). Initial conditions ensure that our solution fits exactly into the unique requirements of the problem.

Exponential functions are fundamental in the study of differential equations. They frequently appear as solutions because of their unique properties.
These functions can grow rapidly and are especially useful in modeling patterns of growth or decay, like population growth or radioactive decay.
In the context of our problem, the exponential function is given as \( e^{x^2/2} \). It's important due to its behavior during differentiation. Differentiating an exponential function of the form \( e^{u(x)} \) requires using the chain rule.

• The derivative of \( e^{x^2/2} \) is \( x e^{x^2/2} \) because the derivative of the power \( x^2/2 \) with respect to \( x \) is \( x \).

We see how the exponential function influences the behavior of our solution \( y(x) = A + B e^{x^2/2} \).
By understanding this, we can adeptly manipulate and solve the differential equations involving such expressions.

Derivatives provide critical insight into the behavior of functions, especially within differential equations. They measure how a quantity changes as another quantity changes.
In our differential equation \( y' - xy - x = 0 \), the term \( y' \) represents the derivative of \( y \) with respect to \( x \).
The exercise required substituting the function \( y(x) = A + B e^{x^2/2} \) into the differential equation. Therefore, we needed to compute its derivative.

• The derivative of \( A + B e^{x^2/2} \) is \( Bx e^{x^2/2} \).
• This uses the chain rule since the inner function \( x^2/2 \) has a simple derivative \( x \).
• With these derivatives, we substituted back into the equation, allowing us to simplify and solve for \( A \).

Understanding derivatives enables us to manipulate equations that describe changing systems, solving them accurately.
This skill is foundational in any calculus or differential equations coursework.