Most Important Selected Qs for JEE AdvancedMathematicsDifferential Equations
If the given curve satisfies the differential equation e^y d x+ (x e^y+2 y ) d y=0 and also passes through (0,0) then the possible equation of curve can be
Options
- Ax e^y+y=0
- Bx+y^ e^y=0
- Cx^2 e^x+y e^y=1
- Dnone of these
Correct answer
D. none of these
Step-by-step solution
M=e^y, N=x e^y+2 y Now, dM dy (keeping x constant )= e ^ y = dN dx (keeping y constant) Hence, the given equation is exact Now, f= e^y d x+g(y)=x e^y+g(y) Differentiating w.r.t y we get, x e^y+ d g d y =N=x e^y+2 y i.e. d g d y =2 y Integrating g(y)=y^2+c₁ f = x e ^y+ y ^2+ c ₁ So, desired solution is x e ^y+ y ^2=0