Let x (d y/d x)-y=x2(x ex+ex-1) ∀ x ∈ R- 0 such that y(1)=e-1. If y(2)=k y(1)(y(1)+2), then the value of k is not less than

Question

Let x (d y/d x)-y=x2(x ex+ex-1) ∀ x ∈ R- 0 such that y(1)=e-1. If y(2)=k y(1)(y(1)+2), then the value of k is not less than (A) 3 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

Θ x (d y/d x)-y=x2(x ex+ex-1) ⇒ ( dy / dx )-(1/ x ) ⋅ y = x ( xe x + e x -1) text I.F. =e∫ (-1/x) d x=e- ln x=(1/x) ∴ Solution is y y ⋅ (1/x)=∫(x ex+ex-1) d x=x ex-x+c Θ y(1) =e-1 ⇒ e-1=e-1+c ⇒ c=0 ∴ y=x2 (ex-1) ∴ y(2) =4(e2-1) =4(e-1)(e+1) =4 y(1)(y(1)+2) ∴ k =4.

Q. Let xdxdy​−y=x2(xex+ex−1)∀x∈R−{0} such that y(1)=e−1. If y(2)=k y(1)(y(1)+2), then the value of k is not less than

3

4

5

6

Θxdxdy​−y=x2(xex+ex−1) ⇒dxdy​−x1​⋅y=x(xex+ex−1) I.F. =e∫x−1​dx=e−lnx=x1​ ∴ Solution is yy⋅x1​=∫(xex+ex−1)dx=xex−x+c Θy(1)=e−1⇒e−1=e−1+c⇒c=0 ∴y=x2(ex−1) ∴y(2)=4(e2−1) =4(e−1)(e+1) =4y(1)(y(1)+2) ∴k=4.

Q. Let $x \frac{d y}{d x} - y = x^{2} (x e^{x} + e^{x} - 1) \forall x \in R - {0}$ such that $y (1) = e - 1$ . If $y (2) = k$ $y (1) (y (1) + 2)$ , then the value of $k$ is not less than