Find the values of A and B so that the function y(x)=Ae×+Bxe×+x²e× satisfy the initial conditions y(1)=1, y'(1)= -1

To find the values of \( A \) and \( B \) such that the function \( y(x) = Ae^x + Bxe^x + x^2e^x \) satisfies the initial conditions \( y(1) = 1 \) and \( y'(1) = -1 \), follow these steps:

1. **Find \( y(1) = 1 \):**
   
   Substitute \( x = 1 \) into \( y(x) \):
   \[
   y(1) = A e^1 + B(1)e^1 + (1)^2 e^1 = A e + B e + e
   \]
   \[
   y(1) = e(A + B + 1)
   \]
   Since \( y(1) = 1 \):
   \[
   e(A + B + 1) = 1 \quad \Rightarrow \quad A + B + 1 = \frac{1}{e}
   \]
   \[
   A + B = \frac{1}{e} - 1 \quad \text{(Equation 1)}
   \]

2. **Find \( y'(x) \):**
   
   Differentiate \( y(x) \):
   \[
   y'(x) = \frac{d}{dx} (Ae^x + Bxe^x + x^2e^x)
   \]
   Using the product rule:
   \[
   y'(x) = A e^x + B(e^x + xe^x) + (2xe^x + x^2e^x)
   \]
   \[
   y'(x) = A e^x + B e^x + Bxe^x + 2xe^x + x^2e^x
   \]
   \[
   y'(x) = (A + B)e^x + (B + 2)xe^x + x^2e^x
   \]

3. **Find \( y'(1) = -1 \):**
   
   Substitute \( x = 1 \) into \( y'(x) \):
   \[
   y'(1) = (A + B)e + (B + 2)(1)e + (1)^2e
   \]
   \[
   y'(1) = (A + B)e + (B + 2)e + e
   \]
   \[
   y'(1) = e(A + B) + e(B + 2) + e
   \]
   \[
   y'(1) = e(A + B + B + 2 + 1) = e(A + 2B + 3)
   \]
   Since \( y'(1) = -1 \):
   \[
   e(A + 2B + 3) = -1 \quad \Rightarrow \quad A + 2B + 3 = -\frac{1}{e}
   \]
   \[
   A + 2B = -\frac{1}{e} - 3 \quad \text{(Equation 2)}
   \]

4. **Solve the system of equations:**

   From Equation 1:
   \[
   A + B = \frac{1}{e} - 1
   \]

   From Equation 2:
   \[
   A + 2B = -\frac{1}{e} - 3
   \]

   Subtract Equation 1 from Equation 2:
   \[
   (A + 2B) - (A + B) = \left(-\frac{1}{e} - 3\right) - \left(\frac{1}{e} - 1\right)
   \]
   \[
   B = -\frac{2}{e} - 2
   \]

   Substitute \( B \) back into Equation 1:
   \[
   A + \left(-\frac{2}{e} - 2\right) = \frac{1}{e} - 1
   \]
   \[
   A - \frac{2}{e} - 2 = \frac{1}{e} - 1
   \]
   \[
   A = \frac{1}{e} - 1 + \frac{2}{e} + 2
   \]
   \[
   A = \frac{3}{e} + 1
   \]

Thus, the values of \( A \) and \( B \) are:
\[
A = \frac{3}{e} + 1
\]
\[
B = -\frac{2}{e} - 2
\]

Do you want details or have any questions?

Here are some related questions you might be interested in:

1. How do you find the general solution to a first-order differential equation?
2. What is the method for solving systems of linear equations?
3. How do you differentiate a product of two functions?
4. What are the common techniques for solving initial value problems?
5. How do you apply the chain rule in differentiation?
6. What is the importance of exponential functions in calculus?
7. How do you interpret the solution of a differential equation graphically?
8. What is the relationship between differential equations and their initial conditions?

**Tip:** Always verify your solution by substituting back into the original equations to ensure consistency.

Learn how to find the values of A and B for the function y(x)=Ae^x+Bxe^x+x^2e^x that satisfy the initial conditions y(1)=1 and y'(1)=-1. This problem involves differential equations and exponential functions, with detailed step-by-step solutions using differentiation rules and the product rule.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation