solve the following differential equuation .(1+y^2)(1+logx)dx+xdy=0

To solve the differential equation \((1+y^2)(1+\log x)dx + xdy = 0\), we can start by rearranging the terms to isolate \(dy\) and \(dx\). This will help us identify if the equation can be separated or if we can use an integrating factor. Let's break it down step by step.

Rearranging the Equation

We can rewrite the equation in a more manageable form:

(1 + y²)(1 + log x)dx + xdy = 0

Rearranging gives us:

xdy = -(1 + y²)(1 + log x)dx

Now, we can express this as:

dy/dx = -((1 + y²)(1 + log x))/x

Separation of Variables

Next, we can separate the variables \(y\) and \(x\). This means we want all terms involving \(y\) on one side and all terms involving \(x\) on the other side:

dy / (1 + y²) = -((1 + log x)/x)dx

Integrating Both Sides

Now we can integrate both sides. The left side involves the integral of \(1/(1 + y²)\), which is a standard integral:

• ∫(1/(1 + y²)) dy = arctan(y) + C₁

For the right side, we can simplify the integral:

• ∫-((1 + log x)/x) dx = -∫(1/x) dx - ∫(log x/x) dx
• ∫(1/x) dx = log x
• For ∫(log x/x) dx, we can use integration by parts:

Let u = log x and dv = (1/x)dx. Then, du = (1/x)dx and v = x. Applying integration by parts:

• ∫log x dx = x log x - ∫x(1/x)dx = x log x - x + C₂

Putting it all together, we have:

arctan(y) = -log x - (x log x - x) + C

Final Form of the Solution

We can simplify this further:

arctan(y) = -log x - x log x + x + C

To express \(y\) explicitly, we can take the tangent of both sides:

y = tan(-log x - x log x + x + C)

Summary

The solution to the differential equation \((1+y^2)(1+\log x)dx + xdy = 0\) is:

y = tan(-log x - x log x + x + C)

where \(C\) is the constant of integration. This solution captures the relationship between \(y\) and \(x\) as dictated by the original differential equation.