MHT CET202124 Sep 2021Evening ShiftMathematicsDifferential EquationsActual
The particular solution of the differential equation (1+e^ 2 x ) d y+e^x (1+y^2 ) d x=0 at x=0 and y=1 is
Options
- A⁻¹ e ^ x - ⁻¹ y =0
- B⁻¹ e ^ x + ⁻¹ y = 2
- C⁻¹ e ^ x + ⁻¹ y = 3 4
- D⁻¹ e^x- ⁻¹ y= 3 4
Correct answer
B. ⁻¹ e ^ x + ⁻¹ y = 2
Step-by-step solution
aligned & (1+ e ^ 2 x ) dy + e ^ x (1+ y ^2 ) dx =0 & dy 1+ y ^2 + e ^ x (1+ e ^ 2 x ) dx =0 & dy 1+ y ^2 =- e ^ x 1+ e ^ 2 x d x aligned Put e ^ x = t e ^ x dx = dt dy 1+ y ^2 =- dt 1+ t ^2 ⁻¹( y )=- ⁻¹( t )+ c ⁻¹( y )+ ⁻¹ ( e ^ x )= c We have x =0, y =1 ⁻¹(1)+ ⁻¹ ( e ^ )= c c =2 ⁻¹(1)= 2 ⁻¹ y + ⁻¹ e ^ x = 2