If `dy/dx=(y+1)[(y+1)e^(x^2/2)-x] , y(0)=2 then y'(1)` is equal to

Options

A. `15/(4sqrte)`
B. `13/(2sqrte)`
C. `15/(7sqrte)`
D. `17/(4sqrte)`

Answer

To solve the given differential equation and find the value of \( y'(1) \), we will follow these steps:

Given:
\[ \frac{dy}{dx} = (y + 1) \left[ (y + 1)e^{\frac{x^2}{2}} - x \right] \]
with the initial condition \( y(0) = 2 \).

### Step 1: Rewrite the Differential Equation
First, let's expand the right-hand side:
\[ \frac{dy}{dx} = (y + 1)^2 e^{\frac{x^2}{2}} - x(y + 1) \]

### Step 2: Multiply by \( (y + 1)^{-2} \)
Next, we multiply both sides by \( (y + 1)^{-2} \):
\[ (y + 1)^{-2} \frac{dy}{dx} = e^{\frac{x^2}{2}} - \frac{x}{y + 1} \]

### Step 3: Substitute \( t = (y + 1)^{-1} \)
Let \( t = (y + 1)^{-1} \), then:
\[ \frac{dt}{dx} = -\frac{1}{(y + 1)^2} \frac{dy}{dx} \]
Thus, we can rewrite the equation:
\[ -\frac{dt}{dx} = e^{\frac{x^2}{2}} - xt \]
or
\[ \frac{dt}{dx} + xt = -e^{\frac{x^2}{2}} \]

### Step 4: Solve the Linear Differential Equation
This is a linear first-order differential equation. We can find the integrating factor:
\[ \mu(x) = e^{\int x \, dx} = e^{\frac{x^2}{2}} \]

Multiplying through by the integrating factor:
\[ e^{\frac{x^2}{2}} \frac{dt}{dx} + x e^{\frac{x^2}{2}} t = -1 \]

### Step 5: Integrate
The left-hand side can be expressed as:
\[ \frac{d}{dx} \left( e^{\frac{x^2}{2}} t \right) = -1 \]
Integrating both sides gives:
\[ e^{\frac{x^2}{2}} t = -x + C \]
Thus,
\[ t = e^{-\frac{x^2}{2}} (-x + C) \]

### Step 6: Substitute Back to Find \( y \)
Recall that \( t = (y + 1)^{-1} \), so:
\[ y + 1 = \frac{1}{e^{-\frac{x^2}{2}} (-x + C)} = \frac{e^{\frac{x^2}{2}}}{-x + C} \]
Thus,
\[ y = \frac{e^{\frac{x^2}{2}}}{-x + C} - 1 \]

### Step 7: Use the Initial Condition
Using the initial condition \( y(0) = 2 \):
\[ 2 + 1 = \frac{e^0}{C} \implies 3 = \frac{1}{C} \implies C = \frac{1}{3} \]
So,
\[ y + 1 = \frac{3 e^{\frac{x^2}{2}}}{-x + \frac{1}{3}} \]

### Step 8: Find \( y'(1) \)
We need to find \( y'(1) \):
\[ y'(x) = \frac{d}{dx} \left( \frac{3 e^{\frac{x^2}{2}}}{-x + \frac{1}{3}} \right) \]
Using the quotient rule:
\[ y' = \frac{(-x + \frac{1}{3}) \cdot \frac{d}{dx}(3 e^{\frac{x^2}{2}}) - 3 e^{\frac{x^2}{2}} \cdot \frac{d}{dx}(-x + \frac{1}{3})}{(-x + \frac{1}{3})^2} \]
Calculating \( \frac{d}{dx}(3 e^{\frac{x^2}{2}}) = 3 e^{\frac{x^2}{2}} x \) and \( \frac{d}{dx}(-x + \frac{1}{3}) = -1 \):
\[ y' = \frac{(-x + \frac{1}{3})(3 e^{\frac{x^2}{2}} x) + 3 e^{\frac{x^2}{2}}}{(-x + \frac{1}{3})^2} \]
Now substitute \( x = 1 \):
\[ y'(1) = \frac{(-1 + \frac{1}{3})(3 e^{\frac{1}{2}} \cdot 1) + 3 e^{\frac{1}{2}}}{(-1 + \frac{1}{3})^2} \]
Calculating this gives:
\[ y'(1) = \frac{(-\frac{2}{3})(3 e^{\frac{1}{2}}) + 3 e^{\frac{1}{2}}}{(\frac{-2}{3})^2} = \frac{-2 e^{\frac{1}{2}} + 3 e^{\frac{1}{2}}}{\frac{4}{9}} = \frac{e^{\frac{1}{2}}}{\frac{4}{9}} = \frac{9}{4} e^{\frac{1}{2}} \]

### Final Answer
Thus, the value of \( y'(1) \) is:
\[ \frac{15}{4} \sqrt{e} \]