Question Number 69790 by mathmax by abdo last updated on 27/Sep/19
$${sove}\:\left({x}^{\mathrm{2}} −\mathrm{3}{x}\right){y}^{''} \:\:+\mathrm{2}{x}\:{y}^{'} \:=\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{−{x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 01/Oct/19
![we put y^′ =z so (e) ⇔(x^2 −3x)z^′ +2xz =(2x+1)e^(−x^2 ) (e) (he) →(x^2 −3x)z^′ +2x z =0 ⇒(x^2 −3x)z^′ =−2xz ⇒ (z^′ /z)=((−2x)/(x^2 −3x)) =((−2)/(x−3)) ⇒ln∣z∣=−2ln∣x−3∣ +α ⇒ z =(K/(∣x−3∣^2 )) let determine the solution on ]3,+∞[ z =(K/((x−3)^2 )) mvc method →z^′ =(K^′ /((x−3)^2 )) +K×((−2(x−3)^ )/((x−3)^4 )) =(K^′ /((x−3)^2 )) −((2K)/((x−3)^3 )) (e) ⇒(x^2 −3x){(K^′ /((x−3)^2 ))−((2K)/((x−3)^3 ))}+2x(K/((x−3)^2 )) =(2x+1)e^(−x^2 ) ⇒ (x/(x−3))K^′ −2(x/((x−3)^2 ))K +((2xK)/((x−3)^2 )) =(2x+1)e^(−x^2 ) ⇒ K^′ =(((2x+1)(x−3))/x) e^(−x^2 ) ⇒K^′ =((2x^2 −6x+x−3)/x)e^(−x^2 ) ⇒ K^′ =(((2x^2 −5x−3))/x) e^(−x^2 ) ⇒K(x) =∫^x (((2t^2 −5t−3))/t) e^(−t^2 ) dt +C ⇒ z(x)=(1/((x−3)^2 )){ ∫^x (((2t^2 −5t−3))/t)e^(−t^2 ) dt +C} y^′ (x)=z(x) ⇒y(x) =∫z(x)dx z is known](https://www.tinkutara.com/question/Q70144.png)
$${we}\:{put}\:{y}^{'} ={z}\:\:{so}\:\left({e}\right)\:\Leftrightarrow\left({x}^{\mathrm{2}} −\mathrm{3}{x}\right){z}^{'} \:+\mathrm{2}{xz}\:=\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{−{x}^{\mathrm{2}} } \:\:\:\left({e}\right) \\ $$$$\left({he}\right)\:\rightarrow\left({x}^{\mathrm{2}} −\mathrm{3}{x}\right){z}^{'} \:+\mathrm{2}{x}\:{z}\:=\mathrm{0}\:\:\:\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{3}{x}\right){z}^{'} =−\mathrm{2}{xz}\:\Rightarrow \\ $$$$\frac{{z}^{'} }{{z}}=\frac{−\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{3}{x}}\:=\frac{−\mathrm{2}}{{x}−\mathrm{3}}\:\Rightarrow{ln}\mid{z}\mid=−\mathrm{2}{ln}\mid{x}−\mathrm{3}\mid\:+\alpha\:\Rightarrow \\ $$$$\left.{z}\:=\frac{{K}}{\mid{x}−\mathrm{3}\mid^{\mathrm{2}} }\:\:\:\:{let}\:{determine}\:{the}\:{solution}\:{on}\:\right]\mathrm{3},+\infty\left[\right. \\ $$$${z}\:=\frac{{K}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:\:{mvc}\:{method}\:\rightarrow{z}^{'} =\frac{{K}^{'} }{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:+{K}×\frac{−\mathrm{2}\left({x}−\mathrm{3}\right)^{} }{\left({x}−\mathrm{3}\right)^{\mathrm{4}} } \\ $$$$=\frac{{K}^{'} }{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:−\frac{\mathrm{2}{K}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$\left({e}\right)\:\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{3}{x}\right)\left\{\frac{{K}^{'} }{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{K}}{\left({x}−\mathrm{3}\right)^{\mathrm{3}} }\right\}+\mathrm{2}{x}\frac{{K}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:=\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{−{x}^{\mathrm{2}} } \:\Rightarrow \\ $$$$\frac{{x}}{{x}−\mathrm{3}}{K}^{'} \:\:−\mathrm{2}\frac{{x}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }{K}\:\:+\frac{\mathrm{2}{xK}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\:=\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{−{x}^{\mathrm{2}} } \:\Rightarrow \\ $$$${K}^{'} \:=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left({x}−\mathrm{3}\right)}{{x}}\:{e}^{−{x}^{\mathrm{2}} } \:\Rightarrow{K}^{'} \:=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}+{x}−\mathrm{3}}{{x}}{e}^{−{x}^{\mathrm{2}} } \:\Rightarrow \\ $$$${K}^{'} \:=\frac{\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}\right)}{{x}}\:{e}^{−{x}^{\mathrm{2}} \:} \Rightarrow{K}\left({x}\right)\:=\int^{{x}} \frac{\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{5}{t}−\mathrm{3}\right)}{{t}}\:{e}^{−{t}^{\mathrm{2}} } {dt}\:+{C}\:\Rightarrow \\ $$$${z}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{3}\right)^{\mathrm{2}} }\left\{\:\int^{{x}} \:\frac{\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{5}{t}−\mathrm{3}\right)}{{t}}{e}^{−{t}^{\mathrm{2}} } {dt}\:+{C}\right\} \\ $$$${y}^{'} \left({x}\right)={z}\left({x}\right)\:\Rightarrow{y}\left({x}\right)\:=\int{z}\left({x}\right){dx}\:\:\:{z}\:{is}\:{known} \\ $$