KVPY2015MathematicsDifferential Equations
Let f: R R be a continuous function satisfying f(x)+ ₀^ x t f(t) d t+x²=0 for all x R . Then
Options
- A_ x f(x)=2
- B_ x - f(x)=-2
- Cf(x) has more than one point in common with the x -axis
- Df(x) is an odd function
Correct answer
B. _ x - f(x)=-2
Step-by-step solution
f^ (x)+x f(x)+2 x=0 d y d x +x y=-2 x I.F. =e^ x² / 2 y e^ x² / 2 = e^ x² / 2 (-2 x) d x+Ce^ x² / 2 y=-2 e^ x² / 2 +Cy=-2+C e^ i x² / 2 f(0)=0 C=2f(x)=2 [e^ -x² / 2 -1 ]