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

Question

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

Emma Aldridge · Accepted Answer

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

You have found the differential equation - it is in the question! We assume you seek the solution of the differential equation and also that the log base is #e#. # (1+y^2)(1+lnx)dx+xdy=0# with #y(1)=1# ..... [A] We can rearrange this ODE from differential form to standard and collect terms: # - 1/(1+y^2)dy/dx = (1+lnx)/x # This now separable, so separating the variables yields: # \ \ \ \ \ - \ int \ 1/(1+y^2) \ dy = int \ (1+lnx)/x \ dx # # :. - \ int \ 1/(1+y^2) \ dy = int \ 1/x \ dx + int \ lnx/x \ dx # ..... [B] The first and and second integrals are standard, the third will require an application of integration by parts: Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x, => v,=lnx ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # gives us # int \ (lnx)(1/x) \ dx = (lnx)(lnx) - int \ (lnx)(1/x) \ dx # # :. 2 \ int \ (lnx)/x \ dx = ln^2x => int \ (lnx)/x \ dx = ln^2x/2 # Using this result, we can now return to integrating our earlier result [B]: # - arctany = lnx + (ln^2x)/2 + C # Applying the initial condition #y(1)=1# we have: # - arctan1 = ln1 + (ln^2 1)/2 + C => C = -pi/4# Thus: # - arctany = lnx + (ln^2x)/2 - pi/4 # # :. arctany = pi/4 - lnx - ln^2x # # :. y = tan(pi/4 - lnx - ln^2x) #

Ezekiel Alexander · Answer

undefined 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:

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