The order of the differential equation of family of curves y= tan ((π/4)+a x) tan ((π/4)-a x)+c eb x+d is (where a, b, c, d are parameters)

Question

The order of the differential equation of family of curves y= tan ((π/4)+a x) tan ((π/4)-a x)+c eb x+d is (where a, b, c, d are parameters) (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

y=(1+ tan a x/1- tan a x) ⋅ (1- tan a x/1+ tan a x)+( operatornameced) eb x=1+k eb x two arbitary constant ∴ order 2

Q. The order of the differential equation of family of curves $y = tan (\frac{π}{4} + a x) tan (\frac{π}{4} - a x) + c e^{b x + d}$ is (where a, b, c, d are parameters)

1

2

3

4

$y = \frac{1 + t a n a x}{1 - t a n a x} \cdot \frac{1 - t a n a x}{1 + t a n a x} + (ce^{d}) e^{b x} = 1 + k e^{b x}$
two arbitary constant
$∴$ order 2

Q. The order of the differential equation of family of curves y=tan(4π​+ax)tan(4π​−ax)+cebx+d is (where a, b, c, d are parameters)

1

2

3

4

y=1−tanax1+tanax​⋅1+tanax1−tanax​+(ced)ebx=1+kebx two arbitary constant ∴ order 2

Q. The order of the differential equation of family of curves $y = tan (\frac{π}{4} + a x) tan (\frac{π}{4} - a x) + c e^{b x + d}$ is (where a, b, c, d are parameters)

$y = \frac{1 + t a n a x}{1 - t a n a x} \cdot \frac{1 - t a n a x}{1 + t a n a x} + (ce^{d}) e^{b x} = 1 + k e^{b x}$
two arbitary constant
$∴$ order 2