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 50431 by Abdo msup. last updated on 16/Dec/18

solve xy^′ +y =((2x)/(√(1−x^4 )))

$${solve}\:{xy}^{'} \:+{y}\:=\frac{\mathrm{2}{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }} \\ $$

Commented by Abdo msup. last updated on 16/Dec/18

(he) xy^′ +y =0 ⇒xy^′ =−y ⇒(y^′ /y)=−(1/x) ⇒ln∣y∣=−ln∣x∣ +k ⇒ y(x)=(K/(∣x∣)) let take x>0⇒y=(K/x) mvc method give y^′ =(K^′ /x) −(K/x^2 ) and (e) ⇔ K^′ −(K/x) +(K/x) =((2x)/(√(1−x^4 ))) ⇒ K^′ = ((2x)/(√(1−x^4 ))) ⇒ K(x)= ∫ ((2xdx)/(√(1−x^4 ))) changement x^2 =sint give K(x) =∫ ((cost dt)/(√(1−sin^2 t))) =∫ dt +c =t +λ =arcsin(x^2 ) +λ ⇒ y(x)= ((arcsin(x^2 ) +λ)/x) =(λ/x) + ((arcsin(x^2 ))/x) .

$$\left({he}\right)\:\:\:\:\:{xy}^{'} \:+{y}\:=\mathrm{0}\:\Rightarrow{xy}^{'} =−{y}\:\Rightarrow\frac{{y}^{'} }{{y}}=−\frac{\mathrm{1}}{{x}}\:\Rightarrow{ln}\mid{y}\mid=−{ln}\mid{x}\mid\:+{k}\:\Rightarrow \\ $$$${y}\left({x}\right)=\frac{{K}}{\mid{x}\mid}\:\:\:{let}\:{take}\:{x}>\mathrm{0}\Rightarrow{y}=\frac{{K}}{{x}}\:\:{mvc}\:{method}\:{give}\: \\ $$$${y}^{'} =\frac{{K}^{'} }{{x}}\:−\frac{{K}}{{x}^{\mathrm{2}} }\:\:{and}\:\left({e}\right)\:\Leftrightarrow\:\:{K}^{'} \:−\frac{{K}}{{x}}\:+\frac{{K}}{{x}}\:=\frac{\mathrm{2}{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\Rightarrow \\ $$$${K}^{'} \:=\:\frac{\mathrm{2}{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\Rightarrow\:{K}\left({x}\right)=\:\int\:\:\:\frac{\mathrm{2}{xdx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:\:{changement}\:{x}^{\mathrm{2}} ={sint} \\ $$$${give}\:{K}\left({x}\right)\:=\int\:\:\frac{{cost}\:{dt}}{\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}}\:=\int\:{dt}\:+{c}\:={t}\:+\lambda\:\:={arcsin}\left({x}^{\mathrm{2}} \right)\:+\lambda \\ $$$$\Rightarrow\:{y}\left({x}\right)=\:\frac{{arcsin}\left({x}^{\mathrm{2}} \right)\:+\lambda}{{x}}\:=\frac{\lambda}{{x}}\:+\:\frac{{arcsin}\left({x}^{\mathrm{2}} \right)}{{x}}\:. \\ $$$$ \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18

xdy+ydx=((2x)/(√(1−(x^2 )^2 )))dx d(xy)=((d(x^2 ))/(√(1−(x^2 )^2 ))) ∫d(xy)=∫((d(x^2 ))/(√(1−(x^2 )^2 ))) xy=sin^(−1) (x^2 )+c

$${xdy}+{ydx}=\frac{\mathrm{2}{x}}{\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }}{dx} \\ $$$${d}\left({xy}\right)=\frac{{d}\left({x}^{\mathrm{2}} \right)}{\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\int{d}\left({xy}\right)=\int\frac{{d}\left({x}^{\mathrm{2}} \right)}{\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$${xy}={sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{c} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com