find-f-x-if-f-x-f-x-2-2x-1-

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

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} \dots . .

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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