Obtain the general solution:
[1 + xy]dx = (x^2 + 1)dy

y=x+ 0.5c(1+ x^2)

y = x + c(1 + x^2)^0.5

y=x+ c(1+ x^2)

y= 0.5x+ 0.5c(1+ x^2)

We are tasked with solving the differential equation:

\[
(1 + xy)dx = (x^2 + 1)dy
\]

### Step 1: Rearrange into a more workable form.
We rewrite the given equation as:

\[
\frac{dy}{dx} = \frac{1 + xy}{x^2 + 1}
\]

### Step 2: Use a substitution.
We can try the substitution \( y = vx \), where \( v = \frac{y}{x} \). Then:

\[
y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}
\]

Substituting these into the original equation gives:

\[
v + x\frac{dv}{dx} = \frac{1 + x(vx)}{x^2 + 1} = \frac{1 + x^2v}{x^2 + 1}
\]

### Step 3: Simplify the equation.
Now, simplify and isolate the derivative:

\[
v + x\frac{dv}{dx} = \frac{1 + x^2v}{x^2 + 1}
\]

Subtract \( v \) from both sides:

\[
x\frac{dv}{dx} = \frac{1 + x^2v}{x^2 + 1} - v
\]

Express everything under a common denominator on the right side:

\[
x\frac{dv}{dx} = \frac{1 + x^2v - v(x^2 + 1)}{x^2 + 1}
\]

This simplifies to:

\[
x\frac{dv}{dx} = \frac{1 + x^2v - vx^2 - v}{x^2 + 1} = \frac{1 - v}{x^2 + 1}
\]

### Step 4: Solve the resulting separable equation.
Now, we have a separable equation:

\[
\frac{dv}{1 - v} = \frac{dx}{x^2 + 1}
\]

### Step 5: Integrate both sides.
We integrate both sides:

\[
\int \frac{dv}{1 - v} = \int \frac{dx}{x^2 + 1}
\]

The integrals are straightforward:

\[
-\ln|1 - v| = \tan^{-1}(x) + C
\]

Exponentiate both sides:

\[
1 - v = e^{-\tan^{-1}(x) - C}
\]

We can express the constant \( e^{-C} \) as another constant \( k \):

\[
1 - v = \frac{k}{\sqrt{1 + x^2}}
\]

### Step 6: Substitute back for \( v \).
Recall that \( v = \frac{y}{x} \), so:

\[
1 - \frac{y}{x} = \frac{k}{\sqrt{1 + x^2}}
\]

Multiply by \( x \):

\[
x - y = k\frac{x}{\sqrt{1 + x^2}}
\]

Thus, the general solution is:

\[
y = x + \frac{kx}{\sqrt{1 + x^2}}
\]

### Conclusion:
Among the given options, the general solution corresponds to:

\[
y = x + c(1 + x^2)^{0.5}
\]

which matches the option:

\[
y = x + c(1 + x^2)^{0.5}
\]

Would you like more details on any step or have any questions?

Here are five related questions to consider:
1. How does the substitution \( y = vx \) help simplify the equation?
2. What techniques are used to solve separable differential equations?
3. How do you handle constants of integration during differential equation solutions?
4. What role does trigonometric substitution play in solving integrals like \( \int \frac{dx}{x^2 + 1} \)?
5. How can you verify the solution to a differential equation?

**Tip:** When solving differential equations, always check if a substitution simplifies the variables and makes the equation separable.

Obtain the general solution: [1 + xy]dx = (x^2 + 1)dy

This page provides a detailed solution to the differential equation [1 + xy]dx = (x^2 + 1)dy. Using the substitution method and solving a separable equation, the solution involves key mathematical concepts such as differential equations and integrals. Suitable for advanced high school or undergraduate students.

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