Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 102323 by Learner101 last updated on 08/Jul/20

Solve this linear Equation (dy/dx) + y cos x = (1/2) sin x

$${Solve}\:{this}\:{linear}\:{Equation} \\ $$$$\frac{{dy}}{{dx}}\:+\:{y}\:{cos}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\:{x} \\ $$

Answered by bemath last updated on 08/Jul/20

IF u(x)=e^(∫cos x dx) = e^(sin x) y(x)=((∫ (1/2)sin x e^(sin x) dx +C)/e^(sin x) ) y(x)= (((1/2)∫sin xe^(sin x) dx +C)/e^(sin x) ) y(x)= e^(−sin x) {(1/2)∫sin x e^(sin x) +C}

$${IF}\:\:{u}\left({x}\right)={e}^{\int\mathrm{cos}\:{x}\:{dx}} \:=\:{e}^{\mathrm{sin}\:{x}} \\ $$$${y}\left({x}\right)=\frac{\int\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}\:{e}^{\mathrm{sin}\:{x}} \:{dx}\:+{C}}{{e}^{\mathrm{sin}\:{x}} } \\ $$$${y}\left({x}\right)=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:{xe}^{\mathrm{sin}\:{x}} \:{dx}\:+{C}}{{e}^{\mathrm{sin}\:{x}} } \\ $$$${y}\left({x}\right)=\:{e}^{−\mathrm{sin}\:{x}} \left\{\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:{x}\:{e}^{\mathrm{sin}\:{x}} \:+{C}\right\} \\ $$

Commented by bemath last updated on 08/Jul/20

yes sir.

$${yes}\:{sir}.\: \\ $$

Answered by mathmax by abdo last updated on 08/Jul/20

y^′ +y cosx =((sinx)/2) (he)→y^′ =−ycosx ⇒(y^′ /y) =−cosx ⇒ln∣y∣ =−sinx +c ⇒y =k e^(−sinx) lagrange method →y^′ =k^′ e^(−sinx) −cosx k e^(−sinx) e ⇒k^′ e^(−sinx) −kcosx e^(−sinx) +kcosx e^(−sinx) =(1/2)sinx ⇒ k^′ e^(−sinx) =(1/2)sinx ⇒ k^′ =(1/2)sinx e^(sinx) ⇒ k =(1/2)∫^x sint e^(sint) dt +c ⇒ y(x) =((1/2)∫^x sint e^(sint) dt +c)e^(−sinx) =c e^(−sinx ) +(e^(−sinx) /2) ∫^x sint e^(sint) dt

$$\mathrm{y}^{'} \:+\mathrm{y}\:\mathrm{cosx}\:=\frac{\mathrm{sinx}}{\mathrm{2}} \\ $$$$\left(\mathrm{he}\right)\rightarrow\mathrm{y}^{'} \:=−\mathrm{ycosx}\:\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}\:=−\mathrm{cosx}\:\Rightarrow\mathrm{ln}\mid\mathrm{y}\mid\:=−\mathrm{sinx}\:+\mathrm{c}\:\Rightarrow\mathrm{y}\:=\mathrm{k}\:\mathrm{e}^{−\mathrm{sinx}} \\ $$$$\mathrm{lagrange}\:\mathrm{method}\:\rightarrow\mathrm{y}^{'} \:=\mathrm{k}^{'} \:\mathrm{e}^{−\mathrm{sinx}} \:−\mathrm{cosx}\:\mathrm{k}\:\mathrm{e}^{−\mathrm{sinx}} \\ $$$$\mathrm{e}\:\Rightarrow\mathrm{k}^{'} \:\mathrm{e}^{−\mathrm{sinx}} −\mathrm{kcosx}\:\mathrm{e}^{−\mathrm{sinx}} \:+\mathrm{kcosx}\:\mathrm{e}^{−\mathrm{sinx}} \:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinx}\:\Rightarrow \\ $$$$\mathrm{k}^{'} \:\mathrm{e}^{−\mathrm{sinx}} \:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinx}\:\Rightarrow\:\mathrm{k}^{'} \:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinx}\:\mathrm{e}^{\mathrm{sinx}} \:\Rightarrow\:\mathrm{k}\:=\frac{\mathrm{1}}{\mathrm{2}}\int^{\mathrm{x}} \:\mathrm{sint}\:\mathrm{e}^{\mathrm{sint}} \:\mathrm{dt}\:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{2}}\int^{\mathrm{x}} \:\mathrm{sint}\:\mathrm{e}^{\mathrm{sint}} \:\mathrm{dt}\:+\mathrm{c}\right)\mathrm{e}^{−\mathrm{sinx}} \\ $$$$=\mathrm{c}\:\mathrm{e}^{−\mathrm{sinx}\:} \:+\frac{\mathrm{e}^{−\mathrm{sinx}} }{\mathrm{2}}\:\int^{\mathrm{x}} \mathrm{sint}\:\mathrm{e}^{\mathrm{sint}} \:\mathrm{dt} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com