Question

x(1+p²) = 1

Please solve this

{ Differential Equation }

( Solvable for x, y or z}​

Answer 1

Answer :

x = 1/(1 + p²)

y = p/(1 + p²) - arctan(p) + c

Solution :

Here ,

The given differntial equation is ;

x(1 + p²) = 1 , where p = dy/dx

The given differntial equation can be rewritten as ;

x = 1/(1 + p²) --------(1)

Clearly ,

eq-(1) is solvable for x .

Now ,

Differentiating eq-(1) with respect to y ,

We have ;

=> dx/dy = [-2p/(1 + p²)²]•(dp/dy)

=> 1/p = [-2p/(1 + p²)²]•(dp/dy)

=> dy = [-2p²/(1 + p²)²]•dp

Now ,

Let p = tan∅ , then

dp = sec²∅•d∅

Thus ,

=> dy = [-2tan²∅/(1 + tan²∅)²]•sec²∅•d∅

=> dy = [-2tan²∅/(sec²∅)²]•sec²∅•d∅

=> dy = [-2tan²∅/sec⁴∅]•sec²∅•d∅

=> dy = [-2tan²∅/sec²∅]•d∅

=> dy = -2sin²∅•d∅

Also ,

We know that cos2∅ = 1 - 2sin²∅ , thus

-2sin²∅ = cos2∅ - 1

Thus ,

=> dy = (cos2∅ - 1)•d∅

=> dy = cos2∅•d∅ - d∅

Now ,

Integrating both the sides , we get ;

=> y = ½•sin2∅ - ∅ + c

=> y = ½•[2tan∅/(1 + tan²∅)] - arctan(p) + c

=> y = p/(1 + p²) - arctan(p) + c , where c is an arbitrary constant -------(2)

Thus ,

eq-(1) and eq-(2) together form the general solution in parametric form with p as parameter .