P panky Jul 2011 408 16 Nov 9, 2019 #1 If a function \(\displaystyle f\) satisfies the relation \(\displaystyle f(x)f''(x)-f(x)f'(x)=(f'(x))^2\,\forall x \in\mathbb{R}\) and \(\displaystyle f(0)=f'(0)=1\), then value of \(\displaystyle f(x)\) is . Last edited by a moderator: Nov 9, 2019
If a function \(\displaystyle f\) satisfies the relation \(\displaystyle f(x)f''(x)-f(x)f'(x)=(f'(x))^2\,\forall x \in\mathbb{R}\) and \(\displaystyle f(0)=f'(0)=1\), then value of \(\displaystyle f(x)\) is .
idontknow Dec 2015 972 128 Earth Nov 9, 2019 #2 \(\displaystyle \displaystyle y = x^{2} +x+1\) Last edited: Nov 9, 2019
skipjack Forum Staff Dec 2006 21,379 2,410 Nov 9, 2019 #3 That's incorrect. $f(x) = e^{{\large e^{\large x}} - 1}$ Reactions: 3 people
P panky Jul 2011 408 16 Nov 11, 2019 #4 skipjack said: That's incorrect. $f(x) = e^{{\large e^{\large x}} - 1}$ Click to expand... Yes, skipjack. How do I solve it? Plaase explain to me. Last edited by a moderator: Nov 11, 2019
skipjack said: That's incorrect. $f(x) = e^{{\large e^{\large x}} - 1}$ Click to expand... Yes, skipjack. How do I solve it? Plaase explain to me.
skipjack Forum Staff Dec 2006 21,379 2,410 Nov 11, 2019 #5 Make the substitution $f(x) = e^{\large v}$, then solving is easy. Reactions: panky