verify-that-y-x-e-x-cos-e-x-e-x-sin-e-x-is-the-solution-of-integral-equation-y-x-1-xe-2x-cos-1-e-2x-sin-1-0-x-1-x-t-e-2x-y-t-dt-

Question Number 74218 by malikmasood3535@gmail.com last updated on 20/Nov/19

v e r i f y t h a t y (x) = e^{x} (\cos e^{x} - e^{x} \sin e^{x}) i s t h e s o l u t i o n o f i n t e g r a l e q u a t i o n y (x) = (1 - {x e}^{2 x}) \cos 1 - e^{2 x} \sin 1 + \underset{0}{\int^{x}} {1 - (x - t) e^{2 x}} y (t) d t

Answered by mind is power last updated on 20/Nov/19

y (x) = (1 - {x e}^{2 x}) c o s (1) - e^{2 x} s i n (1) + \int_{0}^{x} (1 - (x - t) e^{2 x}} y (t) d t

y (0) = c o s (1) - s i n (1)

g (x) = \int_{0}^{x} (1 - (x - t) e^{2 x} y (t) d t

g (x) = (1 - {x e}^{2 x}) \int_{0}^{x} y (t) d t + e^{2 x} \int_{0}^{x} t y (t) d t

f o r y (t) = e^{t} (c o s (e^{t}) - e^{t} s o n (e^{t}))

\int_{0}^{x} e^{t} (c o s (e^{t})) - e^{2 t} s i n (e^{t}) d t = e^{x} c o s (e^{x}) - c o s (1)

\int_{0}^{x} t . (e^{t} (c o s (e^{t}) - e^{t} s i n (e^{t})) d t b y p a r t

= {[t . e^{t} c o s (e^{t})]}_{0}^{x} - \int e^{t} c o s (e^{t}) d t

= {x e}^{x} c o s (e^{x}) - [{s i n e}^{t}] = {x e}^{x} c o s (e^{x}) - {s i n e}^{x} + s i n (1)

g (x) = (1 - {x e}^{2 x}) (e^{x} c o s (e^{x}) - c o s (1)) + e^{2 x} ({x e}^{x} c o s (e^{x}) - s i n (e^{x}) + s i n (1))

= - c o s (1) + {x e}^{2 x} c o s (1) + e^{x} c o s (e^{x}) - e^{2 x} s i n (e^{x}) + e^{2 x} s i n (1)

(1 - {x e}^{2 x}) c o s (1) - e^{2 x} s i n (1) + g (x)

= e^{x} c o s (e^{x}) - e^{2 x} s i n (e^{x}) = y (x)

\Rightarrow y (x) e^{x} (c o s (e^{x}) - e^{x} s i n (e^{x})) i s s o l u t i o n