Question Number 94328 by i jagooll last updated on 18/May/20
![y′ + xy = x](https://www.tinkutara.com/question/Q94328.png)
$$\mathrm{y}'\:+\:\mathrm{xy}\:=\:\mathrm{x}\: \\ $$
Answered by i jagooll last updated on 18/May/20
![](https://www.tinkutara.com/question/13116.png)
Commented by i jagooll last updated on 18/May/20
![it correct ?](https://www.tinkutara.com/question/Q94330.png)
$$\mathrm{it}\:\mathrm{correct}\:? \\ $$
Answered by abdomathmax last updated on 18/May/20
![(he)→y^′ +xy =0→(y^′ /y)=−x ⇒ln∣y∣ =−(x^2 /2) +λ ⇒ y = k e^(−(x^2 /2)) let use mvc method(lagrange) y^′ =k^′ e^(−(x^2 /2)) −xk e^(−(x^2 /2)) (e)⇒k^′ e^(−(x^2 /2)) −kx e^(−(x^2 /2)) +kx e^(−(x^2 /2)) =x ⇒ k^′ =x e^(x^2 /2) ⇒k =∫ x e^(x^2 /2) +c =e^(x^2 /2) +c ⇒ y(x) =(e^(x^2 /2) +c)e^(−(x^2 /2)) ⇒y(x) = 1+c e^(−(x^2 /2))](https://www.tinkutara.com/question/Q94443.png)
$$\left(\mathrm{he}\right)\rightarrow\mathrm{y}^{'} +\mathrm{xy}\:=\mathrm{0}\rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}=−\mathrm{x}\:\Rightarrow\mathrm{ln}\mid\mathrm{y}\mid\:=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\lambda\:\Rightarrow \\ $$$$\mathrm{y}\:=\:\mathrm{k}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:\:\:\:\mathrm{let}\:\mathrm{use}\:\mathrm{mvc}\:\mathrm{method}\left(\mathrm{lagrange}\right) \\ $$$$\mathrm{y}^{'} \:=\mathrm{k}^{'} \:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} −\mathrm{xk}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\left(\mathrm{e}\right)\Rightarrow\mathrm{k}^{'} \:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:−\mathrm{kx}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\mathrm{kx}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:=\mathrm{x}\:\Rightarrow \\ $$$$\mathrm{k}^{'} \:=\mathrm{x}\:\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\Rightarrow\mathrm{k}\:=\int\:\mathrm{x}\:\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\mathrm{c}\:=\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\mathrm{c}\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\left(\mathrm{e}^{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\mathrm{c}\right)\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)\:=\:\mathrm{1}+\mathrm{c}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$