find f(x) if f ′(x) + f(x^2 ) = 2x+1


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Question Number 85236 by jagoll last updated on 20/Mar/20

$find f (x) if$ $f^{'} (x) + f (x^{2}) = 2 x + 1$
Commented by mathmax by abdo last updated on 20/Mar/20
$its clear that f is polynomial let f(x)=Σ_(n=0) ^∞ a_n x^n f^′ (x) =Σ_(n=1) ^∞ na_n x^(n−1) =Σ_(n=0) ^∞ (n+1)a_(n+1) x^n f(x^2 ) =Σ_(n=0) ^∞ a_n x^(2n) ⇒ Σ_(n=0) ^∞ (n+1)a_(n+1) x^n +Σ_(n=0) ^∞ a_n x^(2n) =2x+1 changement of indice 2n =p ⇒Σ_(n=0) ^∞ (n+1)a_n x^n +Σ_(p=0) ^∞ a_([(p/2)]) x^p =2x+1 ⇒ Σ_(n=0) ^∞ {(n+1)a_n +a_([(n/2)]) }x^n =2x+1 ⇒ { ((a_0 +a_0 =1)),((2a_1 +a_0 =2 and (n+1)a_n +a_([(n/2)]) =0 ∀n≥2 ⇒)) :} { ((a_0 =(1/2) and a_n =−(a_([(n/2)]) /(n+1)) ∀n≥2)),((a_1 =(3/4))) :} a_2 =−(a_1 /3) , a_3 =−(a_1 /4) .....$
$i t s c l e a r t h a t f i s p o l y n o m i a l l e t f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ $f^{^{'}} (x) = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1} = \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n}$ $f (x^{2}) = \sum_{n = 0}^{\infty} a_{n} x^{2 n} \Rightarrow$ $\sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} + \sum_{n = 0}^{\infty} a_{n} x^{2 n} = 2 x + 1 c h a n g e m e n t o f$ $i n d i c e 2 n = p \Rightarrow \sum_{n = 0}^{\infty} (n + 1) a_{n} x^{n} + \sum_{p = 0}^{\infty} a_{[\frac{p}{2}]} x^{p} = 2 x + 1 \Rightarrow$ $\sum_{n = 0}^{\infty} {(n + 1) a_{n} + a_{[\frac{n}{2}]}} x^{n} = 2 x + 1 \Rightarrow$ ${\begin{cases} a_{0} + a_{0} = 1 \\ 2 a_{1} + a_{0} = 2 a n d (n + 1) a_{n} + a_{[\frac{n}{2}]} = 0 \forall n ⩾ 2 \Rightarrow \end{cases}$ ${\begin{cases} a_{0} = \frac{1}{2} a n d a_{n} = - \frac{a_{[\frac{n}{2}]}}{n + 1} \forall n ⩾ 2 \\ a_{1} = \frac{3}{4} \end{cases}$ $a_{2} = - \frac{a_{1}}{3}, a_{3} = - \frac{a_{1}}{4} . . . . .$
	Answered by john santu last updated on 20/Mar/20

	$\frac{df}{dx} = 2 x + 1 - f (x^{2})$ $df = (2 x + 1) dx - f (x^{2}) dx$ $\int df = \int (2 x + 1) dx - \int f (x^{2}) dx$ $f (x) = x^{2} + x - \int f (x^{2}) dx$ $K = \int f (x^{2}) dx$ $let u = x^{2} \Rightarrow du = 2 x dx$ $dx = \frac{du}{2 \sqrt{u}}$ $K = \int f (u) \times \frac{2 du}{\sqrt{u}}$
	Commented by jagoll last updated on 20/Mar/20

	$how to solve \int f (u) \times \frac{2 d u}{\sqrt{u}}$
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