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Question Number 101846 by bemath last updated on 05/Jul/20

Answered by bramlex last updated on 05/Jul/20

⇔ (dy/dx) −2x^(−1) y = x^2 cos x integrating factor u = e^(∫−2x^(−1) dx) u = e^(−2 ln(x)) = x^(−2) ⇔ y = ((∫ x^(−2) .x^2 cos x dx)/x^(−2) ) ⇔ y = x^2 { sin x + C } y = x^2 sin x + Cx^2 (Bram−□ )

$$\Leftrightarrow\:\frac{{dy}}{{dx}}\:−\mathrm{2}{x}^{−\mathrm{1}} {y}\:=\:{x}^{\mathrm{2}} \:\mathrm{cos}\:{x} \\ $$$${integrating}\:{factor}\:\:{u}\:=\:{e}^{\int−\mathrm{2}{x}^{−\mathrm{1}} \:{dx}} \\ $$$${u}\:=\:{e}\:^{−\mathrm{2}\:\mathrm{ln}\left({x}\right)} \:=\:{x}^{−\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{y}\:=\:\frac{\int\:{x}^{−\mathrm{2}} .{x}^{\mathrm{2}} \mathrm{cos}\:{x}\:{dx}}{{x}^{−\mathrm{2}} } \\ $$$$\Leftrightarrow\:\mathrm{y}\:=\:{x}^{\mathrm{2}} \:\left\{\:\mathrm{sin}\:{x}\:+\:{C}\:\right\}\: \\ $$$$\mathrm{y}\:=\:{x}^{\mathrm{2}} \:\mathrm{sin}\:{x}\:+\:{Cx}^{\mathrm{2}} \: \\ $$$$\left({B}\mathrm{ram}−\square\:\right)\: \\ $$

Answered by mathmax by abdo last updated on 05/Jul/20

(1/x)y^′ −(2/x^2 )y =xcosx ⇒xy^′ −2y =x^3 cosx (x>0) he →xy^′ −2y =0 ⇒xy^′ =2y ⇒(y^′ /y) =(2/x) ⇒ln∣y∣ =2lnx +c ⇒ y =k x^2 lsgrange method →y^′ =k^′ x^2 +2kx e ⇒k^′ x^3 +2kx^2 −2kx^2 =x^3 cosx ⇒k^′ =cosx ⇒k =sinx +λ ⇒ y(x) =(sinx +λ)x^2 =λx^2 +x^2 sinx

$$\frac{\mathrm{1}}{\mathrm{x}}\mathrm{y}^{'} −\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\mathrm{y}\:=\mathrm{xcosx}\:\Rightarrow\mathrm{xy}^{'} \:−\mathrm{2y}\:=\mathrm{x}^{\mathrm{3}} \:\mathrm{cosx}\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}>\mathrm{0}\right) \\ $$$$\mathrm{he}\:\rightarrow\mathrm{xy}^{'} −\mathrm{2y}\:=\mathrm{0}\:\Rightarrow\mathrm{xy}^{'} \:=\mathrm{2y}\:\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}\:=\frac{\mathrm{2}}{\mathrm{x}}\:\Rightarrow\mathrm{ln}\mid\mathrm{y}\mid\:=\mathrm{2lnx}\:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{y}\:=\mathrm{k}\:\mathrm{x}^{\mathrm{2}} \:\:\:\:\mathrm{lsgrange}\:\mathrm{method}\:\rightarrow\mathrm{y}^{'} \:=\mathrm{k}^{'} \mathrm{x}^{\mathrm{2}} \:+\mathrm{2kx} \\ $$$$\mathrm{e}\:\Rightarrow\mathrm{k}^{'} \:\mathrm{x}^{\mathrm{3}} \:+\mathrm{2kx}^{\mathrm{2}} \:−\mathrm{2kx}^{\mathrm{2}} \:=\mathrm{x}^{\mathrm{3}} \:\mathrm{cosx}\:\Rightarrow\mathrm{k}^{'} \:=\mathrm{cosx}\:\Rightarrow\mathrm{k}\:=\mathrm{sinx}\:+\lambda\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\left(\mathrm{sinx}\:+\lambda\right)\mathrm{x}^{\mathrm{2}} \:=\lambda\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}^{\mathrm{2}} \:\mathrm{sinx} \\ $$

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