99 Percentile Qs Bank for JEE MainMathematicsDifferential Equations
Solve of the differential equation ( d y d x = 1+y^2 ( ⁻¹ y )-x )
Options
- A(x e^ ⁻¹ y =e^ - ⁻¹ y ( ( ⁻¹ y )-1 )+c )
- B(x e^ ⁻¹ y =e^ ⁻¹ y ( ( ⁻¹ y )-1 )+c )
- C(x e^ ⁻¹ y =e^ ⁻¹ y ( ( ⁻¹ y )+1 )+c )
- D(x e^ ⁻¹ y =e^ - ⁻¹ y ( ( ⁻¹ y )+1 )+c )
Correct answer
B. (x e^ ⁻¹ y =e^ ⁻¹ y ( ( ⁻¹ y )-1 )+c )
Step-by-step solution
Given differential equation ( d y d x = 1+y^2 ⁻¹ y-x d x d y = ⁻¹ y-x 1+y^2 ) ( d x d y + x 1+y^2 = ⁻¹ y 1+y^2 ) is a linear differential equation, so. ( IF =e^ d y 1+y^2 =e^ ⁻¹ y ) So, the solution is ( aligned & x e^ ⁻¹ y = e^ ⁻¹ y ⁻¹ y 1+y^2 d y+c x e^ ⁻¹ y & = t e^t d t+c, where t= ⁻¹ y x e^ ⁻¹ y & =t e^t- e^t+c x e^ ⁻¹ y & = ( ⁻¹ y ) e^ ⁻¹ y -e^ ⁻¹ y +c x e^ ⁻¹ y & =e^ ⁻¹ y ( ( ⁻¹ y )-1 )+c aligned ) Hence, option (b) is correct.