MHT CET202019 Oct 2020Evening ShiftMathematicsDifferential EquationsActual
The general solution of the differential equation (1+y² )+ (x-e^ ⁻¹ y ) d y d x =0 is
Options
- Ax e^ ⁻¹ y = (e^ ⁻¹ x )² 2 +c
- Be^ ⁻¹ y = (e^ ⁻¹ x )²+c
- Cx e^ ⁻¹ y = (e^ t a n⁻¹ y )² 2 +c
- De^ ⁻¹ y = (e^ ⁻¹ y )²+c
Correct answer
C. x e^ ⁻¹ y = (e^ t a n⁻¹ y )² 2 +c
Step-by-step solution
(A) array l (1+y² )+ (x-e^ ⁻¹ y ) d y d x =0 (x-e^ ⁻¹ y ) d y d x =- (1+y² ) d y d x = - (1+y² ) x-e^ ⁻¹ y d x d y = (x-e^ ⁻¹ y ) - (1+y² ) d x d y = -x (1+y² ) + e^ ⁻¹ y 1+y² d x d y + x 1+y² = e^ ⁻¹ y 1+y² I.F. =e^ 1 1+y² d y =e^ ⁻¹ y array So, the general solution is array l x e^ ⁻¹ y = e^ ⁻¹ y 1+y² e^ ⁻¹ y d x Put e^ ⁻¹ y =t e^ ⁻¹ y 1+y² d y=d t x e^ ⁻¹ y = t d t x e^ ⁻¹ y = t² 2 +c x e^ ⁻¹ y = (e^ ⁻¹ y )² 2 +c array