7. A function y=f(x) satisfies the differential equation (x+1) f^' (x)-2(x^2+x)f(x)=e^(x^2 )/((x+1) ); ∀ x>-1.If f(0)=5,thenf(x)is:

Given the differential equation:

\[
(x+1) f'(x) - 2(x^2 + x) f(x) = \frac{e^{x^2}}{x+1}
\]

with the initial condition \( f(0) = 5 \), we need to solve for the function \( f(x) \).

### Step 1: Rewrite the equation
Let's first rewrite the differential equation in a standard linear form:

\[
f'(x) - \frac{2(x^2 + x)}{x+1} f(x) = \frac{e^{x^2}}{(x+1)^2}
\]

Here, the differential equation is linear in \( f(x) \) with the integrating factor approach being a suitable method to solve it.

### Step 2: Identify the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) \, dx}
\]

where \( P(x) = -\frac{2(x^2 + x)}{x+1} \).

Let's compute \( \int P(x) \, dx \):

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

First, simplify the integrand:

\[
\frac{2(x^2 + x)}{x+1} = 2x \cdot \frac{x+1}{x+1} = 2x
\]

So the integral becomes:

\[
\int -2x \, dx = -x^2
\]

Thus, the integrating factor \( \mu(x) \) is:

\[
\mu(x) = e^{-x^2}
\]

### Step 3: Multiply the differential equation by the integrating factor
Multiplying the entire differential equation by \( e^{-x^2} \):

\[
e^{-x^2} f'(x) - 2x e^{-x^2} f(x) = \frac{e^{x^2}}{(x+1)^2} \cdot e^{-x^2}
\]

Simplifying this:

\[
e^{-x^2} f'(x) - 2x e^{-x^2} f(x) = \frac{1}{(x+1)^2}
\]

Notice that the left-hand side is the derivative of \( e^{-x^2} f(x) \):

\[
\frac{d}{dx} \left( e^{-x^2} f(x) \right) = \frac{1}{(x+1)^2}
\]

### Step 4: Integrate both sides
Integrate both sides with respect to \( x \):

\[
e^{-x^2} f(x) = \int \frac{1}{(x+1)^2} \, dx
\]

The integral on the right-hand side can be found as:

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

So:

\[
e^{-x^2} f(x) = -\frac{1}{x+1} + C
\]

### Step 5: Solve for \( f(x) \)
Now, solve for \( f(x) \):

\[
f(x) = e^{x^2} \left( -\frac{1}{x+1} + C \right)
\]

### Step 6: Apply the initial condition
Use the initial condition \( f(0) = 5 \) to determine \( C \):

\[
f(0) = e^{0} \left( -\frac{1}{1} + C \right) = -1 + C = 5
\]

Thus:

\[
C = 6
\]

### Final Solution
Substitute \( C = 6 \) back into the expression for \( f(x) \):

\[
f(x) = e^{x^2} \left( \frac{6x + 5}{x+1} \right)
\]

This is the solution for \( f(x) \).

Would you like more details or have any questions? 

Here are some related questions for further exploration:
1. How do you find the general solution of a linear first-order differential equation?
2. What is the importance of the integrating factor in solving differential equations?
3. Can you solve a similar differential equation with a different initial condition?
4. How would the solution change if the initial condition were \( f(1) = 3 \)?
5. What happens if the differential equation is nonlinear? How would the approach differ?
6. Can you explain the role of boundary conditions in differential equations?
7. How do you interpret the solution of this differential equation in a real-world context?
8. Can this method be extended to second-order differential equations?

**Tip:** When solving differential equations, always check if the equation can be simplified before applying a method, as it might save considerable time.

Learn how to solve the differential equation (x+1) f'(x) - 2(x^2 + x) f(x) = e^(x^2)/(x+1) with initial condition f(0) = 5. Detailed step-by-step solution involving integrating factor method and initial value problem. Suitable for undergraduate level students.

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