Give the complete solution of this problem make sure your answer is correct. 1. \( x^{2} \frac{d y}{d x}-y(1-x)=0 \quad y(-1)=-1 \)

Let's solve the differential equation:

\[
x^2 \frac{dy}{dx} - y(1-x) = 0, \quad y(-1) = -1.
\]

### Step 1: Simplify the Differential Equation

We can rewrite the equation in a more standard form:

\[
\frac{dy}{dx} = \frac{y(1-x)}{x^2}.
\]

This is a first-order linear differential equation.

### Step 2: Find the Integrating Factor

To solve this differential equation, we need to find the integrating factor. The standard form of the equation is:

\[
\frac{dy}{dx} + P(x)y = Q(x),
\]

where \(P(x) = -\frac{1-x}{x^2}\) and \(Q(x) = 0\).

The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1-x}{x^2} \, dx}.
\]

Let's compute this integral:

\[
\int -\frac{1-x}{x^2} \, dx = \int \left(-\frac{1}{x^2} + \frac{x}{x^2}\right) \, dx = \int \left(-\frac{1}{x^2} + \frac{1}{x}\right) \, dx.
\]

\[
\int \left(-\frac{1}{x^2} + \frac{1}{x}\right) \, dx = \frac{1}{x} + \ln|x|.
\]

Thus, the integrating factor is:

\[
\mu(x) = e^{\frac{1}{x} + \ln|x|} = \frac{e^{\frac{1}{x}}}{x}.
\]

### Step 3: Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor \( \mu(x) = \frac{e^{\frac{1}{x}}}{x} \):

\[
\frac{e^{\frac{1}{x}}}{x} \frac{dy}{dx} + \frac{e^{\frac{1}{x}}}{x} \left(-\frac{1-x}{x^2}\right) y = 0.
\]

This simplifies to:

\[
\frac{d}{dx} \left(\frac{y e^{\frac{1}{x}}}{x}\right) = 0.
\]

### Step 4: Integrate Both Sides

Integrating both sides with respect to \(x\):

\[
\frac{y e^{\frac{1}{x}}}{x} = C,
\]

where \( C \) is the constant of integration.

### Step 5: Solve for \(y(x)\)

Now, solve for \( y(x) \):

\[
y e^{\frac{1}{x}} = Cx,
\]

\[
y = Cx e^{-\frac{1}{x}}.
\]

### Step 6: Apply the Initial Condition

Use the initial condition \( y(-1) = -1 \) to determine \( C \):

\[
-1 = C(-1) e^{-\frac{1}{-1}} = -C e^{-1}.
\]

Thus,

\[
C = e.
\]

### Final Solution

Substitute \( C = e \) back into the expression for \( y(x) \):

\[
y(x) = e \cdot x \cdot e^{-\frac{1}{x}} = x e^{1-\frac{1}{x}}.
\]

So, the complete solution is:

\[
\boxed{y(x) = x e^{1 - \frac{1}{x}}.}
\]

Would you like further details or have any questions?

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**Relative Questions:**

1. What is the significance of the integrating factor in solving linear differential equations?
2. How do initial conditions affect the solution of a differential equation?
3. Can this method be applied to non-linear differential equations? Why or why not?
4. What are some examples of real-world applications of solving first-order linear differential equations?
5. How would the solution change if the initial condition was different?

**Tip:** Always verify your integrating factor by differentiating it and ensuring it matches the given differential equation’s format.

Explore the complete solution of the differential equation x^2(dy/dx) - y(1-x) = 0 with step-by-step explanations. Learn how to simplify the equation, find the integrating factor, and apply initial conditions to solve for y(x). Suitable for undergraduate students studying differential equations.

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