Question Number 80293 by ~blr237~ last updated on 01/Feb/20
$${Find}\:{all}\:{functions}\:{that}\:\:{satisfy}\:{to}\:\: \\ $$$$\left({E}\right):\:\forall\:{x}\in\mathbb{R}\:\:\:\:\:\:{xf}\left({x}\right)+\int_{\mathrm{0}} ^{{x}} {f}\left({x}−{t}\right){cos}\left(\mathrm{2}{t}\right){dt}={sin}\left(\mathrm{2}{x}\right) \\ $$$$\: \\ $$
Commented by john santu last updated on 02/Feb/20
$$\mathrm{2}{y}+{xy}'\:=\:\mathrm{2cos}\:\mathrm{2}{x}\:\Rightarrow\:{differential} \\ $$$${equation} \\ $$
Commented by ~blr237~ last updated on 02/Feb/20
$${just}\:{check}\:{your}\:{mistake}\:{on}\:{the}\:\left(\mathrm{3}−\mathrm{4}\right){lines}\:\:{crossing} \\ $$
Commented by john santu last updated on 02/Feb/20
![(d/dx)[∫_0 ^x f(u)cos (x−u)du]= f(x)cos (x)+f(0) cos (0) ?](https://www.tinkutara.com/question/Q80308.png)
$$\frac{{d}}{{dx}}\left[\underset{\mathrm{0}} {\overset{{x}} {\int}}{f}\left({u}\right)\mathrm{cos}\:\left({x}−{u}\right){du}\right]= \\ $$$${f}\left({x}\right)\mathrm{cos}\:\left({x}\right)+{f}\left(\mathrm{0}\right)\:\mathrm{cos}\:\left(\mathrm{0}\right)\:? \\ $$
Commented by ~blr237~ last updated on 02/Feb/20
$${No}\:\: \\ $$
Commented by jagoll last updated on 02/Feb/20
$${what}\:{the}\:{right}? \\ $$
Commented by ~blr237~ last updated on 02/Feb/20
$${like}\:{that}….?{what}\:{is}\:{not}\:{valid}?\:{be}\:{clear}?\: \\ $$
Commented by jagoll last updated on 02/Feb/20
$${we}\:{know}\:{that} \\ $$$$\int_{{a}} ^{{x}} \:{f}\left({u}\right){du}\:\Rightarrow\:\frac{{d}}{{dx}}\:\int_{{a}} ^{{x}} {f}\left({u}\right){du}=\:{f}\left({x}\right) \\ $$$${it}\:{clear}\:{by}\:{fundamental} \\ $$$${calculus}\:\mathrm{1}.\: \\ $$$${so}\:\frac{{d}}{{dx}}\int_{\mathrm{1}} ^{\:{x}} {f}\left({u}\right)\mathrm{cos}\:\left({x}−{u}\right){du}\:=\: \\ $$$${f}\left({x}\right)\:\mathrm{cos}\:\left({x}−{x}\right)\:=\:{f}\left({x}\right) \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 02/Feb/20
![sir what is going on? let take f(t)=t F(x)=∫_0 ^x (x−t)cos2tdt by part u′=cos2t and v=x−t F(x)=[(1/2)(x−t)sin2t]_0 ^x +(1/2)∫_0 ^x sin2tdt =(1/2)[((−cos2t)/2)]_0 ^x =((1−cos2x)/4) (dF/dx)=((sin2x)/2) ≠ f(x)=x](https://www.tinkutara.com/question/Q80323.png)
$${sir}\:{what}\:{is}\:{going}\:\:{on}? \\ $$$$\:{let}\:{take}\:{f}\left({t}\right)={t} \\ $$$${F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \left({x}−{t}\right){cos}\mathrm{2}{tdt} \\ $$$$\:\:\:\:\:\:\:{by}\:{part}\:\:\:{u}'={cos}\mathrm{2}{t}\:\:{and}\:{v}={x}−{t} \\ $$$${F}\left({x}\right)=\left[\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{t}\right){sin}\mathrm{2}{t}\right]_{\mathrm{0}} ^{{x}} \:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} {sin}\mathrm{2}{tdt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{−{cos}\mathrm{2}{t}}{\mathrm{2}}\right]_{\mathrm{0}} ^{{x}} =\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{4}} \\ $$$$\frac{{dF}}{{dx}}=\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\:\neq\:{f}\left({x}\right)={x} \\ $$
Commented by jagoll last updated on 02/Feb/20
$${okay}\:{sir} \\ $$
Commented by ~blr237~ last updated on 02/Feb/20
![When you want to derivate F(x)=∫_0 ^(u(x)) f(x,t)dt make sure firstly (by a good stating variable: may be z=(t/(u(x))) ) to remove x at the whole boundary we were having F(x)=∫_0 ^x f(x−t)cos2tdt F(x)=∫_0 ^1 x[f(x−xv)cos(2xv)]dv with v=(t/x) Now you can use (dF/dx)=∫_0 ^1 (∂/∂x) g(x,v)dv where g(x,v)=xf(x−xv)cos(2xv)](https://www.tinkutara.com/question/Q80325.png)
$${When}\:{you}\:{want}\:{to}\:{derivate}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{{u}\left({x}\right)} \:{f}\left({x},{t}\right){dt}\:{make} \\ $$$${sure}\:{firstly}\:\left({by}\:{a}\:{good}\:{stating}\:{variable}:\:{may}\:{be}\:\:{z}=\frac{{t}}{{u}\left({x}\right)}\:\:\right)\:{to}\:{remove}\:{x}\:{at}\:{the}\:{whole}\:{boundary} \\ $$$${we}\:{were}\:{having}\:\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({x}−{t}\right){cos}\mathrm{2}{tdt} \\ $$$${F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left[{f}\left({x}−{xv}\right){cos}\left(\mathrm{2}{xv}\right)\right]{dv}\:\:\:{with}\:\:{v}=\frac{{t}}{{x}}\: \\ $$$${Now}\:{you}\:{can}\:{use}\:\:\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial}{\partial{x}}\:{g}\left({x},{v}\right){dv}\:\:\:\:\:\:\:{where}\:\:{g}\left({x},{v}\right)={xf}\left({x}−{xv}\right){cos}\left(\mathrm{2}{xv}\right) \\ $$
Commented by mr W last updated on 02/Feb/20
$${to}\:{jagoll}\:{sir}: \\ $$$$\int_{{a}} ^{{x}} \:{f}\left({u}\right){du}\:\Rightarrow\:\frac{{d}}{{dx}}\:\int_{{a}} ^{{x}} {f}\left({u}\right){du}=\:{f}\left({x}\right) \\ $$$${is}\:{correct}.\:{but}\:{it}\:{can}\:{not}\:{be}\:{applied}\:{to} \\ $$$$\int_{{a}} ^{{x}} \:{f}\left({u},{x}\right){du}\:\Rightarrow\:\frac{{d}}{{dx}}\:\int_{{a}} ^{{x}} {f}\left({u},{x}\right){du}\neq\:{f}\left({x},{x}\right) \\ $$
Commented by mr W last updated on 02/Feb/20
