# Find the differential equation # (1+y^2)(1+lnx)dx+xdy=0# with #y(1)=1#?

# y = tan(pi/4 - lnx - ln^2x) #

We can rearrange this ODE from differential form to standard and collect terms:

This now separable, so separating the variables yields:

The first and and second integrals are standard, the third will require an application of integration by parts:

Then plugging into the IBP formula:

gives us

Using this result, we can now return to integrating our earlier result [B]:

Thus:

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To find the differential equation from the given expression, rearrange the terms to isolate ( dy ) on one side and ( dx ) on the other side. Then, substitute ( y(1) = 1 ) to determine the value of the constant.

Here's the solution:

[ (1+y^2)(1+\ln(x)) , dx + x , dy = 0 ]

[ (1+y^2)(1+\ln(x)) , dx = - x , dy ]

[ (1+y^2)(1+\ln(x)) , dx = - y , dx ]

[ (1+y^2)(1+\ln(x)) , dx + y , dx = 0 ]

[ dx , (1+y^2)(1+\ln(x) + y) = 0 ]

[ (1+y^2)(1+\ln(x) + y) , dx = 0 ]

Now, ( y(1) = 1 ), so:

[ (1+1^2)(1+\ln(1) + 1) , dx = 0 ]

[ (2)(1+0+1) , dx = 0 ]

[ 4 , dx = 0 ]

[ dx = 0 ]

Thus, the differential equation is:

[ (1+y^2)(1+\ln(x) + y) , dx = 0 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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