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Question Number 28529 by abdo imad last updated on 26/Jan/18
solve the d.e. (x^2 −1)y^′ +xy= x^2 −e^x .
$${solve}\:{the}\:{d}.{e}.\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} +{xy}=\:{x}^{\mathrm{2}} −{e}^{{x}} \:. \\ $$
Commented by abdo imad last updated on 28/Jan/18
h.e. (x^2 −1)y^′ =−xy ⇒ (y^′ /y)= ((−x)/(x^2 −1)) ⇒ln∣y∣= ∫ (x/(1−x^2 ))dx =((−1)/2)ln∣1−x^2 ∣ +c =ln((1/( (√(∣1−x^2 ∣))))) +c ⇒y= (λ/( (√(∣1−x^2 ∣)))) let use mvc method case1 ∣x∣<1 ⇒y= λ(1−x^2 )^(−(1/2)) y^′ = λ^′ (1−x^2 )^((−1)/2) +λx(1−x^2 )^((−3)/(2 )) (e) ⇒ −λ^′ (1−x^2 )^(1/2) −λx (1−x^2 )^((−1)/2) +λx(1−x^2 )^((−1)/2) =x^2 −e^x ⇒ λ^′ (1−x^2 )^(1/2) = e^x −x^2 ⇒ λ^′ = (e^x −x^2 )(1−x^2 )^((−1)/2) ⇒ λ(x) = ∫_. ^x ((e^t −t^2 )/( (√(1−t^2 ))))dt +k and y(x)= (1/( (√(1−x^2 ))))( ∫_. ^x ((e^t −t^2 )/( (√(1−t^2 ))))dt +k) .
$${h}.{e}.\:\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} \:=−{xy}\:\Rightarrow\:\frac{{y}^{'} }{{y}}=\:\frac{−{x}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{ln}\mid{y}\mid=\:\int\:\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:=\frac{−\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}−{x}^{\mathrm{2}} \mid\:\:+{c} \\ $$$$={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}}\right)\:+{c}\:\Rightarrow{y}=\:\frac{\lambda}{\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}}\:{let}\:{use}\:{mvc}\:{method} \\ $$$${case}\mathrm{1}\:\:\:\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{y}=\:\lambda\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${y}^{'} =\:\lambda^{'} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} +\lambda{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{−\mathrm{3}}{\mathrm{2}\:\:}} \:\:\:\:\:\left({e}\right)\:\Rightarrow \\ $$$$−\lambda^{'} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} −\lambda{x}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \:+\lambda{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} ={x}^{\mathrm{2}} \:−{e}^{{x}} \\ $$$$\Rightarrow\:\lambda^{'} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:{e}^{{x}} \:−{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:\lambda^{'} =\:\:\left({e}^{{x}} \:−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \:\:\Rightarrow \\ $$$$\lambda\left({x}\right)\:=\:\:\int_{.} ^{{x}} \:\:\:\frac{{e}^{{t}} \:−{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:\:+{k}\:\:\:{and} \\ $$$${y}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\:\:\:\int_{.} ^{{x}} \:\:\:\frac{{e}^{{t}} \:−{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:\:+{k}\right)\:\:. \\ $$
Commented by abdo imad last updated on 28/Jan/18
for case 2 ∣x∣ >1 we folow the same method.
$${for}\:{case}\:\mathrm{2}\:\:\mid{x}\mid\:>\mathrm{1}\:{we}\:{folow}\:{the}\:{same}\:{method}. \\ $$

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