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Let y(x)=xV(x) be a solution of D.E. x²y''-3xy' 3y=0 if V(0)=0 and V(1)=1, then V(-2) is?
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Let y(x)=xV(x) be a solution of D.E. x²y''-3xy' 3y=0 if V(0)=0 and V(1...
Solution:

Given: y(x) = xV(x), x²y'' - 3xy' + 3y = 0, V(0) = 0, V(1) = 1.

To find: V(-2).

To solve the given differential equation, we can use the method of power series.

1. Power Series Representation:
We assume that the solution y(x) can be represented as a power series in x, i.e., y(x) = Σ aₙxⁿ.

2. Differentiating y(x):
Differentiating y(x) with respect to x, we get:
y'(x) = Σ aₙn xⁿ⁻¹.

3. Differentiating y'(x):
Differentiating y'(x) with respect to x, we get:
y''(x) = Σ aₙn(n-1) xⁿ⁻².

4. Substituting the power series representation in the differential equation:
Substituting the power series representation of y(x), y'(x), and y''(x) in the given differential equation, we get:
x²(Σ aₙn(n-1) xⁿ⁻²) - 3x(Σ aₙn xⁿ⁻¹) + 3(Σ aₙxⁿ) = 0.

5. Simplifying the expression:
Expanding the power series, we get:
Σ aₙn(n-1) xⁿ + Σ -3aₙn xⁿ + Σ 3aₙxⁿ = 0.

Combining the terms with the same power of x, we get:
Σ (aₙn(n-1) - 3aₙn + 3aₙ) xⁿ = 0.

6. Finding the recurrence relation:
For the above equation to hold for all values of x, the coefficients of xⁿ must be zero. Therefore, we have:
aₙn(n-1) - 3aₙn + 3aₙ = 0.

Simplifying the equation, we get:
aₙn(n-1) - 3aₙn + 3aₙ = 0.

7. Solving the recurrence relation:
We can solve this recurrence relation to find the values of aₙ.

8. Finding the solution y(x):
Once we have the values of aₙ, we can substitute them in the power series representation of y(x) to obtain the complete solution.

9. Finding V(-2):
To find V(-2), we substitute x = -2 in the solution y(x) = xV(x). Therefore, we have:
V(-2) = -2V(-2).

10. Conclusion:
Since we don't have the value of V(-2), we cannot determine the exact value of V(-2) without solving the recurrence relation and finding the values of aₙ.
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Let y(x)=xV(x) be a solution of D.E. x²y''-3xy' 3y=0 if V(0)=0 and V(1)=1, then V(-2) is?
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