91Ó°ÊÓ

Derivatives play a pivotal role in calculus and differential equations. They help us understand how a function changes at any point. When we work with simple polynomials like \( y = ax^2 + bx \), finding the derivative involves applying basic rules of differentiation. In this case, the derivative \( y' \) is calculated by imagining how the curve of the graph increases or decreases at each point, depending on the values of \( a \) and \( b \).

• The term \( ax^2 \) differentiates to \( 2ax \) because the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) is applied, multiplying the coefficient by the exponent.
• Similarly, the term \( bx \) reduces to \( b \) as it represents a linear change which is constant along \( x \).

Having found \( y' = 2ax + b \), it sets the stage for verifying if it satisfies the given differential equation. Understanding derivatives is crucial, as they help interpret the behavior of functions, which is essential when solving real-world problems.

Verification is the process of checking whether a particular function satisfies a given differential equation. This involves plugging the solution back into the equation and demonstrating both sides align perfectly.
To verify that \( y = ax^2 + bx \) is a solution for the differential equation \( y' = \frac{y}{x} + ax \), we:

• Substitute \( y \) into the right-hand side of the equation, starting by computing \( \frac{y}{x} \).
• Simplify \( \frac{y}{x} = ax + b \), by cancelling out \( x \) in the numerator and denominator.
• Add \( ax \) to get \( ax + b + ax \), simplifying further to get \( 2ax + b \).

Both sides simplify to the same expression, confirming the function \( y = ax^2 + bx \) satisfies the differential equation. Verification ensures the function behaves correctly under the conditions specified in the differential equation, essentially acting as a double-check of our work.

Step-by-step solutions break down complex problems into manageable parts, making it easier to tackle differential equations. Let’s revisit how we verified the solution for this exercise.
First, differentiate the function \( y = ax^2 + bx \) to get \( y' = 2ax + b \). This is our target for the left-hand side of the differential equation.
Next, substitute \( y = ax^2 + bx \) into the right-hand side \( \frac{y}{x} + ax \). While simplifying \( \frac{y}{x} \), focus on rearranging terms so division is straightforward, i.e., \( \frac{ax^2 + bx}{x} = ax + b \).
Combining like terms gives \( 2ax + b \). This matches the derived \( y' \), showing consistency throughout.

• Breaking the problem into smaller tasks helps avoid errors and simplifies understanding.
• At each step, simplify equations manually to avoid missing essential terms or operations.
• Cross-verify by double-checking obtained results against the calculation steps.

Overall, it is crucial to navigate through each step with clarity. This approach can be applied to other mathematical problems, enhancing problem-solving skills.