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Question Number 90963 by abdomathmax last updated on 27/Apr/20
solve (1+x^2 )y^′ +xy =(√(1+x^2 ))
$${solve}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+{xy}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$
Commented by niroj last updated on 27/Apr/20
(1+x^2 )y^′ +xy = (√(1+x^2 )) (dy/dx)+(x/(1+x^2 ))y= ((√(1+x^2 ))/(1+x^2 )) (dy/dx) + (x/(1+x^2 )) y= (1/( (√(1+x^2 )))) form (dy/dx)+Py=Q let, P= (x/(1+x^2 )) , Q= (1/( (√(1+x^2 )))) IF= e^(∫Pdx) = e^(∫(( x)/(1+x^2 ))dx) Put, x^2 +1=t 2xdx=dt xdx=(dt/2) IF=e^(∫(1/t).(dt/2)) =e^((1/2)log t) =e^(log(√t)) ∴ IF=(√t) =(√(1+x^2 )) Complete solution y.IF= ∫IF.Qdx+c y(√(1+x^2 ))= ∫ (√(1+x^2 )) .(1/( (√(1+x^2 ))))dx+C y(√(1+x^2 )) =∫dx+C y = (1/( (√(1+x^2 ))))(x+C)//.
$$\:\:\:\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}^{'} \:+\mathrm{xy}\:=\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{y}=\:\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{y}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\mathrm{form}\:\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q} \\ $$$$\:\:\mathrm{let},\:\mathrm{P}=\:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:,\:\:\mathrm{Q}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\mathrm{IF}=\:\mathrm{e}^{\int\mathrm{Pdx}} =\:\mathrm{e}^{\int\frac{\:\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}} \\ $$$$\:\mathrm{Put},\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{t} \\ $$$$\:\:\mathrm{2xdx}=\mathrm{dt} \\ $$$$\:\:\:\:\:\:\mathrm{xdx}=\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$\:\:\mathrm{IF}=\mathrm{e}^{\int\frac{\mathrm{1}}{\mathrm{t}}.\frac{\mathrm{dt}}{\mathrm{2}}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{t}} =\mathrm{e}^{\mathrm{log}\sqrt{\mathrm{t}}} \\ $$$$\:\therefore\:\mathrm{IF}=\sqrt{\mathrm{t}}\:\:=\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\mathrm{Complete}\:\mathrm{solution} \\ $$$$\:\:\mathrm{y}.\mathrm{IF}=\:\int\mathrm{IF}.\mathrm{Qdx}+\mathrm{c} \\ $$$$\:\:\mathrm{y}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\:\int\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}+\mathrm{C} \\ $$$$\:\:\mathrm{y}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\int\mathrm{dx}+\mathrm{C} \\ $$$$\:\:\mathrm{y}\:=\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\left(\mathrm{x}+\mathrm{C}\right)//. \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 27/Apr/20
(he)→(1+x^2 )y^′ +xy =0⇒(1+x^2 )y^′ =−xy ⇒(y^′ /y)=−(x/(1+x^2 )) ⇒ ln∣y∣ =−(1/2)ln(1+x^2 )+c ⇒y(x)=(K/( (√(1+x^2 )))) let use mvc method y^′ =(K^′ /( (√(1+x^2 )))) +K(−(1/2)(2x)(1+x^2 )^(−(3/2)) ) =(K^′ /( (√(1+x^2 ))))−Kx(1+x^2 )^(−(3/2)) (e)⇒(√(1+x^2 ))K^′ −((Kx)/( (√(1+x^2 )))) +((xK)/( (√(1+x^2 ))))=(√(1+x^2 )) ⇒ K^′ =1 ⇒K(x) =x+λ ⇒y(x) =((x+λ)/( (√(1+x^2 )))) (λ ∈R)
$$\left({he}\right)\rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+{xy}\:=\mathrm{0}\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:=−{xy}\:\Rightarrow\frac{{y}^{'} }{{y}}=−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{c}\:\Rightarrow{y}\left({x}\right)=\frac{{K}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:{let}\:{use}\:{mvc}\:{method} \\ $$$${y}^{'} \:=\frac{{K}^{'} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{K}\left(−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \right)\:=\frac{{K}^{'} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}−{Kx}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\left({e}\right)\Rightarrow\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{K}^{'} −\frac{{Kx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\frac{{xK}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${K}^{'} =\mathrm{1}\:\Rightarrow{K}\left({x}\right)\:={x}+\lambda\:\Rightarrow{y}\left({x}\right)\:=\frac{{x}+\lambda}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\:\left(\lambda\:\in{R}\right) \\ $$
Answered by john santu last updated on 27/Apr/20
y′ +((x/(1+x^2 )))y = (1/( (√(1+x^2 )))) Integrating factor ⇒∫ p(x) dx = ∫ (x/(1+x^2 )) dx ∫ p(x) dx = (1/2)∫ ((d(1+x^2 ))/(1+x^2 )) = ln (√(1+x^2 )) u(x) = e^(∫ p(x) dx) = (√(1+x^2 )) solution y = ((∫ u(x).q(x) dx +C)/(u(x))) y = ((∫ (√(1+x^2 )) ((1/( (√(1+x^2 ))))) dx+C)/( (√(1+x^2 )))) y = ((x+C)/( (√(1+x^2 )))) .
$${y}'\:+\left(\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){y}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${Integrating}\:{factor} \\ $$$$\Rightarrow\int\:{p}\left({x}\right)\:{dx}\:=\:\int\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\: \\ $$$$\int\:{p}\left({x}\right)\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{d}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\mathrm{ln}\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${u}\left({x}\right)\:=\:{e}^{\int\:{p}\left({x}\right)\:{dx}} \:=\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${solution}\: \\ $$$${y}\:=\:\frac{\int\:{u}\left({x}\right).{q}\left({x}\right)\:{dx}\:+{C}}{{u}\left({x}\right)} \\ $$$${y}\:=\:\frac{\int\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\:{dx}+{C}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${y}\:=\:\frac{{x}+{C}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:.\: \\ $$

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