Let y(x) is solution of differential equation `(y^2 – x) (dy)/(dx)` = 1 and y(0) = 1, then find the value of x where curve cuts the x-axis

Options

A. 1. `-e`
B. 2. `2-e`
C. 3. 2
D. 4. `2+e`

Answer

To solve the differential equation \((y^2 - x) \frac{dy}{dx} = 1\) with the initial condition \(y(0) = 1\), and to find the value of \(x\) where the curve cuts the x-axis, we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the given equation:
\[ (y^2 - x) \frac{dy}{dx} = 1 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{1}{y^2 - x} \]

### Step 2: Separating Variables
We can separate the variables by rewriting the equation:
\[ dx = (y^2 - x) dy \]
This can be rearranged to:
\[ dx + x dy = y^2 dy \]

### Step 3: Identifying the Linear Differential Equation
The equation is now in the form:
\[ \frac{dx}{dy} + x = y^2 \]
This is a linear first-order differential equation.

### Step 4: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:
\[ \mu(y) = e^{\int 1 \, dy} = e^y \]

### Step 5: Multiplying by the Integrating Factor
Multiplying the entire equation by the integrating factor:
\[ e^y \frac{dx}{dy} + e^y x = e^y y^2 \]

### Step 6: Integrating Both Sides
The left-hand side can be expressed as:
\[ \frac{d}{dy}(e^y x) = e^y y^2 \]
Integrating both sides gives:
\[ e^y x = \int e^y y^2 \, dy \]
Using integration by parts, we can solve the integral on the right.

### Step 7: Applying Integration by Parts
Let \(u = y^2\) and \(dv = e^y dy\). Then \(du = 2y dy\) and \(v = e^y\). Applying integration by parts:
\[ \int y^2 e^y \, dy = y^2 e^y - \int 2y e^y \, dy \]
We apply integration by parts again on \(\int 2y e^y \, dy\).

### Step 8: Completing the Integration
After applying integration by parts, we get:
\[ \int y^2 e^y \, dy = y^2 e^y - 2(y e^y - e^y) = y^2 e^y - 2y e^y + 2e^y \]
Thus:
\[ e^y x = y^2 e^y - 2y e^y + 2e^y + C \]

### Step 9: Solving for \(x\)
Dividing by \(e^y\):
\[ x = y^2 - 2y + 2 + Ce^{-y} \]

### Step 10: Applying Initial Condition
Using the initial condition \(y(0) = 1\):
\[ 0 = 1 - 2 + 2 + Ce^{-1} \]
This simplifies to:
\[ 0 = 1 + Ce^{-1} \implies C = -e \]

### Step 11: Final Equation
Substituting \(C\) back into the equation:
\[ x = y^2 - 2y + 2 - e \cdot e^{-y} \]

### Step 12: Finding Where the Curve Cuts the X-axis
The curve cuts the x-axis when \(y = 0\):
\[ x = 0^2 - 2(0) + 2 - e \cdot e^{0} = 2 - e \]

### Final Answer
The value of \(x\) where the curve cuts the x-axis is:
\[ \boxed{2 - e} \]