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To solve this problem using the *reduction of order* method, we start with the given second-order differential equation and a known solution \( y_1(t) = t^2 \):

\[
-5t^2 y'' + 4t y' + 2y = 0.
\]

### Step-by-step solution:

1. **Substitute \( y_2 \) in terms of \( y_1 \):**  
   We assume a second solution of the form:
   \[
   y_2(t) = v(t) y_1(t),
   \]
   where \( v(t) \) is a function to be determined. Since \( y_1(t) = t^2 \), we have:
   \[
   y_2(t) = v(t) t^2.
   \]

2. **Find \( y_2' \) and \( y_2'' \):**  
   Using the product rule:
   \[
   y_2' = v' t^2 + 2vt,
   \]
   \[
   y_2'' = v'' t^2 + 4v' t + 2v.
   \]

3. **Substitute into the original ODE:**  
   Substituting \( y_2, y_2', y_2'' \) into the original ODE:
   \[
   -5t^2 (v'' t^2 + 4v' t + 2v) + 4t (v' t^2 + 2vt) + 2(v t^2) = 0.
   \]
   Simplifying this equation gives:
   \[
   -5t^4 v'' - 20t^3 v' - 10t^2 v + 4t^3 v' + 8t^2 v + 2t^2 v = 0.
   \]
   Combining like terms, this simplifies to:
   \[
   -5t^4 v'' - 16t^3 v' = 0.
   \]

4. **Simplify further and solve for \( v \):**  
   Dividing through by \( -5t^3 \) gives:
   \[
   t v'' + 3 v' = 0.
   \]
   This is a first-order linear equation in terms of \( v' \). Let \( w = v' \), so that \( w' = v'' \). The equation becomes:
   \[
   t w' + 3w = 0.
   \]

5. **Solve for \( w \):**  
   This equation is separable:
   \[
   \frac{w'}{w} = -\frac{3}{t}.
   \]
   Integrating both sides yields:
   \[
   \ln |w| = -3 \ln |t| + C_1 \Rightarrow w = C_1 t^{-3}.
   \]

6. **Find \( v \):**  
   Integrating \( w = v' = C_1 t^{-3} \):
   \[
   v = C_1 \int t^{-3} \, dt = C_1 \left(-\frac{1}{2} t^{-2}\right) + C_2.
   \]

7. **Substitute \( v \) back into \( y_2 \):**  
   Since \( y_2 = v(t) t^2 \):
   \[
   y_2 = \left(C_1 \left(-\frac{1}{2} t^{-2}\right) + C_2\right) t^2.
   \]
   Simplifying, this becomes:
   \[
   y_2 = -\frac{C_1}{2} + C_2 t^2.
   \]

   For a second linearly independent solution, set \( C_2 = 0 \):
   \[
   y_2 = -\frac{1}{2} t^{-1}.
   \]

8. **Apply initial conditions:**  
   The initial conditions are \( y_2(1) = -\frac{5}{11} \) and \( y_2'(1) = \frac{1}{11} \). For the general solution:
   \[
   y_2 = A t^{-1} + B t^2.
   \]

   Substituting \( t = 1 \):
   \[
   A + B = -\frac{5}{11}.
   \]

   For the derivative:
   \[
   y_2' = -A t^{-2} + 2B t.
   \]
   Substituting \( t = 1 \):
   \[
   -A + 2B = \frac{1}{11}.
   \]

9. **Solve for \( A \) and \( B \):**  
   Solving the system of equations:
   \[
   A + B = -\frac{5}{11},
   \]
   \[
   -A + 2B = \frac{1}{11}.
   \]
   Adding these gives:
   \[
   3B = -\frac{4}{11} \Rightarrow B = -\frac{4}{33}.
   \]

   Substituting back:
   \[
   A - \frac{4}{33} = -\frac{5}{11} \Rightarrow A = -\frac{15}{33} + \frac{4}{33} = -\frac{11}{33} = -\frac{1}{3}.
   \]

10. **Final solution for \( y_2 \):**  
   Thus, the second solution is:
   \[
   y_2(t) = -\frac{1}{3} t^{-1} - \frac{4}{33} t^2.
   \]

Would you like more details on any part of the solution or do you have any questions? Here are five related questions:

1. How does the reduction of order method work in general for solving second-order differential equations?
2. What conditions are necessary for using the reduction of order method?
3. How do you determine whether two solutions to an ODE are linearly independent?
4. What are some common alternative methods for solving second-order linear differential equations?
5. How do initial conditions affect the form of the particular solution?

**Tip:** When solving differential equations, always verify if the form of your assumed solution satisfies the initial conditions given.

Given that y1(t) = t^2 is a solution to the ODE: -5t^2y'' + 4ty' + 2y = 0, use the reduction of order method to find another solution y2 corresponding to the initial values y2(1) = -5/11 and y2'(1) = 1/11.

Learn how to apply the reduction of order method to solve the second-order differential equation -5t^2y'' + 4ty' + 2y = 0. This detailed guide includes step-by-step calculations to find the second solution, y2(t), and apply the initial conditions y2(1) = -5/11 and y2'(1) = 1/11. Suitable for university-level students studying differential equations.

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