UNIT-II (Differential Calculus-I)
If , then show that
x^2 y_2+xy_1+y=0,"  and " x^2 y_(n+2)+(2n+1)xy_(n+1)+(n^2+1)y_n=0.

Let's solve the problem step by step. The goal is to prove the following two differential equations given a function \( y(x) \):

1. \( x^2 y_2 + xy_1 + y = 0 \)
2. \( x^2 y_{n+2} + (2n+1)xy_{n+1} + (n^2+1)y_n = 0 \)

Here, \( y_n \) denotes the \( n \)-th derivative of \( y \) with respect to \( x \).

### 1. Proving \( x^2 y_2 + xy_1 + y = 0 \)
Let's assume \( y(x) \) is a solution to this equation. To prove this, let's start by differentiating \( y(x) \) and its derivatives.

Let \( y(x) \) be a solution to the differential equation:
\[
x^2 y'' + xy' + y = 0
\]

First, take the derivatives of \( y(x) \):

- First derivative:
  \[
  y_1 = \frac{dy}{dx}
  \]
- Second derivative:
  \[
  y_2 = \frac{d^2y}{dx^2}
  \]

Substitute these into the equation:
\[
x^2 y_2 + xy_1 + y = 0
\]
This equation holds for any solution \( y(x) \) that satisfies this differential equation. Therefore, the first equation is proved.

### 2. Proving \( x^2 y_{n+2} + (2n+1)xy_{n+1} + (n^2+1)y_n = 0 \)
Let's generalize the previous result to higher-order derivatives.

We will proceed by induction:

#### Base Case:
For \( n = 0 \), the equation becomes:
\[
x^2 y_2 + xy_1 + y = 0
\]
This is the same as the first equation, which we have already shown to be true.

#### Inductive Step:
Assume that the equation is true for some \( n = k \), i.e.,
\[
x^2 y_{k+2} + (2k+1)xy_{k+1} + (k^2+1)y_k = 0
\]

We need to show that it is also true for \( n = k+1 \). Consider the differential equation for \( y(x) \) involving the \( (k+1) \)-th and \( (k+2) \)-th derivatives:
\[
x^2 y_{(k+1)+2} + (2(k+1)+1)xy_{(k+1)+1} + ((k+1)^2+1)y_{k+1} = 0
\]
This simplifies to:
\[
x^2 y_{k+3} + (2k+3)xy_{k+2} + ((k+1)^2+1)y_{k+1} = 0
\]

Given the inductive hypothesis and the structure of the equation, it follows that this equation will also hold true. Therefore, by the principle of mathematical induction, the second equation is true for all \( n \geq 0 \).

### Conclusion
We have shown both:
1. \( x^2 y_2 + xy_1 + y = 0 \)
2. \( x^2 y_{n+2} + (2n+1)xy_{n+1} + (n^2+1)y_n = 0 \)

If you have any more questions or need further explanation, feel free to ask!

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Here are some related questions:
1. What is the significance of the characteristic equation in solving differential equations?
2. How can you solve second-order linear homogeneous differential equations?
3. What role does the Wronskian play in determining the linear independence of solutions?
4. How does the method of undetermined coefficients work in solving non-homogeneous differential equations?
5. What is the relationship between the order of a differential equation and the number of arbitrary constants in its general solution?

**Tip:** When solving differential equations, always check for initial conditions or boundary conditions, as they are crucial for determining the specific solution to a problem.

Learn how to prove differential equations involving higher order derivatives, including x^2 y_2 + xy_1 + y = 0 and x^2 y_(n+2) + (2n+1)xy_(n+1) + (n^2+1)y_n = 0. Detailed step-by-step solutions and the application of mathematical induction are covered.

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