Question Number 74218 by malikmasood3535@gmail.com last updated on 20/Nov/19
$${verify}\:{that}\:{y}\left({x}\right)={e}^{{x}} \left(\mathrm{cos}\:{e}^{{x}} −{e}^{{x}} \mathrm{sin}\:{e}^{{x}} \right)\:{is}\:{the}\:{solution}\:{of}\:{integral}\:{equation}\:{y}\left({x}\right)=\left(\mathrm{1}−{xe}^{\mathrm{2}{x}} \right)\mathrm{cos}\:\mathrm{1}−{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{1}+\underset{\mathrm{0}} {\overset{{x}} {\int}}\left\{\mathrm{1}−\left({x}−{t}\right){e}^{\mathrm{2}{x}} \right\}{y}\left({t}\right){dt} \\ $$
Answered by mind is power last updated on 20/Nov/19
![y(x)=(1−xe^(2x) )cos(1)−e^(2x) sin(1)+∫_0 ^x (1−(x−t)e^(2x) }y(t)dt y(0)=cos(1)−sin(1) g(x)=∫_0 ^x (1−(x−t)e^(2x) y(t)dt g(x)=(1−xe^(2x) )∫_0 ^x y(t)dt+e^(2x) ∫_0 ^x ty(t)dt for y(t)=e^t (cos(e^t )−e^t son(e^t )) ∫_0 ^x e^t (cos(e^t ))−e^(2t) sin(e^t )dt=e^x cos(e^x )−cos(1) ∫_0 ^x t.(e^t (cos(e^t )−e^t sin(e^t ))dt by part =[t.e^t cos(e^t )]_0 ^x −∫e^t cos(e^t )dt =xe^x cos(e^x )−[sine^t ]=xe^x cos(e^x )−sine^x +sin(1) g(x)=(1−xe^(2x) )(e^x cos(e^x )−cos(1))+e^(2x) (xe^x cos(e^x )−sin(e^x )+sin(1)) =−cos(1)+xe^(2x) cos(1)+e^x cos(e^x )−e^(2x) sin(e^x )+e^(2x) sin(1) (1−xe^(2x) )cos(1)−e^(2x) sin(1)+g(x) =e^x cos(e^x )−e^(2x) sin(e^x )=y(x) ⇒y(x)e^x (cos(e^x )−e^x sin(e^x )) is solution](https://www.tinkutara.com/question/Q74230.png)
$${y}\left({x}\right)=\left(\mathrm{1}−{xe}^{\mathrm{2}{x}} \right){cos}\left(\mathrm{1}\right)−{e}^{\mathrm{2}{x}} {sin}\left(\mathrm{1}\right)+\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}−\left({x}−{t}\right){e}^{\mathrm{2}{x}} \right\}{y}\left({t}\right){dt} \\ $$$${y}\left(\mathrm{0}\right)={cos}\left(\mathrm{1}\right)−{sin}\left(\mathrm{1}\right) \\ $$$${g}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}−\left({x}−{t}\right){e}^{\mathrm{2}{x}} {y}\left({t}\right){dt}\right. \\ $$$${g}\left({x}\right)=\left(\mathrm{1}−{xe}^{\mathrm{2}{x}} \right)\int_{\mathrm{0}} ^{{x}} {y}\left({t}\right){dt}+{e}^{\mathrm{2}{x}} \int_{\mathrm{0}} ^{{x}} {ty}\left({t}\right){dt} \\ $$$${for}\:{y}\left({t}\right)={e}^{{t}} \left({cos}\left({e}^{{t}} \right)−{e}^{{t}} {son}\left({e}^{{t}} \right)\right) \\ $$$$\int_{\mathrm{0}} ^{{x}} {e}^{{t}} \left({cos}\left({e}^{{t}} \right)\right)−{e}^{\mathrm{2}{t}} {sin}\left({e}^{{t}} \right){dt}={e}^{{x}} {cos}\left({e}^{{x}} \right)−{cos}\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{{x}} {t}.\left({e}^{{t}} \left({cos}\left({e}^{{t}} \right)−{e}^{{t}} {sin}\left({e}^{{t}} \right)\right){dt}\:\:{by}\:{part}\right. \\ $$$$=\left[{t}.{e}^{{t}} {cos}\left({e}^{{t}} \right)\right]_{\mathrm{0}} ^{{x}} −\int{e}^{{t}} {cos}\left({e}^{{t}} \right){dt} \\ $$$$={xe}^{{x}} {cos}\left({e}^{{x}} \right)−\left[{sine}^{{t}} \right]={xe}^{{x}} {cos}\left({e}^{{x}} \right)−{sine}^{{x}} +{sin}\left(\mathrm{1}\right) \\ $$$${g}\left({x}\right)=\left(\mathrm{1}−{xe}^{\mathrm{2}{x}} \right)\left({e}^{{x}} {cos}\left({e}^{{x}} \right)−{cos}\left(\mathrm{1}\right)\right)+{e}^{\mathrm{2}{x}} \left({xe}^{{x}} {cos}\left({e}^{{x}} \right)−{sin}\left({e}^{{x}} \right)+{sin}\left(\mathrm{1}\right)\right) \\ $$$$=−{cos}\left(\mathrm{1}\right)+{xe}^{\mathrm{2}{x}} {cos}\left(\mathrm{1}\right)+{e}^{{x}} {cos}\left({e}^{{x}} \right)−{e}^{\mathrm{2}{x}} {sin}\left({e}^{{x}} \right)+{e}^{\mathrm{2}{x}} {sin}\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{1}−{xe}^{\mathrm{2}{x}} \right){cos}\left(\mathrm{1}\right)−{e}^{\mathrm{2}{x}} {sin}\left(\mathrm{1}\right)+{g}\left({x}\right) \\ $$$$={e}^{{x}} {cos}\left({e}^{{x}} \right)−{e}^{\mathrm{2}{x}} {sin}\left({e}^{{x}} \right)={y}\left({x}\right) \\ $$$$\Rightarrow{y}\left({x}\right){e}^{{x}} \left({cos}\left({e}^{{x}} \right)−{e}^{{x}} {sin}\left({e}^{{x}} \right)\right)\:{is}\:{solution} \\ $$$$ \\ $$$$ \\ $$