A function y=f(x) satisfies (x+1) ⋅ f prime(x)-2(x2+x) f(x)=(ex2/(x+1)), ∀ x-1 If f(0)=5, then f(x) is

Question

A function y=f(x) satisfies (x+1) ⋅ f prime(x)-2(x2+x) f(x)=(ex2/(x+1)), ∀ x>-1 If f(0)=5, then f(x) is (A) ((3 x +5/ x +1)) ⋅ e x 2 (B) ((6 x+5/x+1)) ⋅ ex2 (C) ((6 x +5/( x +1)2)) ⋅ e x 2 (D) ((5-6 x/x+1)) ⋅ ex2

Tardigrade Admin · Accepted Answer

f prime( x )-(2 x ( x +1)/ x +1) f ( x )=( e x 2/( x +1)2) text I. F. = e ∫-2 xdx = e - x 2 ∴ f ( x ) ⋅ e - x 2=∫ ( dx /( x +1)2) ⇒ f ( x ) ⋅ e - x 2=-(1/ x +1)+ C text at x =0, f (0)=5 ⇒ C =6 ∴ f ( x )=((6 x +5/ x +1)) ⋅ e x 2

Q. A function $y = f (x)$ satisfies $(x + 1) \cdot f^{'} (x) - 2 (x^{2} + x) f (x) = \frac{e ^{x^{2}}}{( x + 1 )}, \forall x > - 1$ If $f (0) = 5$ , then $f (x)$ is

$(\frac{3 x + 5}{x + 1}) \cdot e^{x^{2}}$

$(\frac{6 x + 5}{x + 1}) \cdot e^{x^{2}}$

$(\frac{6 x + 5}{( x + 1 ) ^{2}}) \cdot e^{x^{2}}$

$(\frac{5 - 6 x}{x + 1}) \cdot e^{x^{2}}$

Q. A function y=f(x) satisfies (x+1)⋅f′(x)−2(x2+x)f(x)=(x+1)ex2​,∀x>−1 If f(0)=5, then f(x) is

(x+13x+5​)⋅ex2

(x+16x+5​)⋅ex2

((x+1)26x+5​)⋅ex2

(x+15−6x​)⋅ex2

f′(x)−x+12x(x+1)​f(x)=(x+1)2ex2​ I. F. =e∫−2xdx=e−x2 ∴f(x)⋅e−x2=∫(x+1)2dx​⇒f(x)⋅e−x2=−x+11​+C at x=0,f(0)=5⇒C=6 ∴f(x)=(x+16x+5​)⋅ex2

Q. A function $y = f (x)$ satisfies $(x + 1) \cdot f^{'} (x) - 2 (x^{2} + x) f (x) = \frac{e ^{x^{2}}}{( x + 1 )}, \forall x > - 1$ If $f (0) = 5$ , then $f (x)$ is

$(\frac{3 x + 5}{x + 1}) \cdot e^{x^{2}}$

$(\frac{6 x + 5}{x + 1}) \cdot e^{x^{2}}$

$(\frac{6 x + 5}{( x + 1 ) ^{2}}) \cdot e^{x^{2}}$

$(\frac{5 - 6 x}{x + 1}) \cdot e^{x^{2}}$