My teacher gaves us an excersise and i need help because i cant solve it. It's an antiderivative where i need to calculate the f type.
$$f'(x)=f(x){(x+1)^2\over(x+1)},$$ for $x>0$, $f(0)=1$.
Sorry for my english. Thanks in advance.
Find type of $f(x)$ problem
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$\begingroup$
calculus
ordinary-differential-equations
derivatives
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0Why the $\dfrac{(x+1)^2}{x+1}$ ? – 2017-02-28
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0i don't know. i had that problem too. he said it's better to avoid simplification – 2017-02-28
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0Perhaps he had some special method in mind other than integrating between $0$ and $t$ both sides of the equation $$\frac{f'(x)}{f(x)}=x+1$$ – 2017-02-28
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0Ok. But if f(x)=<0 ? – 2017-02-28
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0@TakisOmegas Why does that matter? – 2017-02-28
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0f.e if f(x)=0 i can't devide with it. I have to find f. So it will be: (ln|f(x)|)'=(x^2/2+x)' ? – 2017-02-28
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0oohhh you have a homogeneous first order differential equation here. Do you know how to solve one of those? – 2017-02-28
1 Answers
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You should write it differently: $$\frac {f'(x)}{f(x)} ={(x+1)^2\over(x+1)}$$ and remember the derivative of $$ log f(x) $$