Question Number 53965 by maxmathsup by imad last updated on 28/May/19
![solve (x+1)y^(′′) +(2−3x^2 )y^′ =xsin(x) let y^′ =z so (e) ⇔(x+1)z^′ +(2−3x^2 )z =xsinx(e) let first find z (he) →(x+1)z^′ +(2−3x^2 )z =0 ⇒(x+1)z^′ =(3x^2 −2)z ⇒(z^′ /z) =((3x^2 −2)/(x+1)) ⇒ ln∣z∣ = ∫ ((3x^2 −2)/(x+1)) dx +c =∫ ((3(x^2 −1)+1)/(x+1))+c =∫3(x−1)dx +∫ (dx/(x+1)) +c =3((x^2 /2)−x)+ln∣x+1∣ +c ⇒z =K∣x+1∣e^((3/2)x^2 −3x) solution on]−1,+∞[ ⇒ z(x) =K(x+1) e^((3/2)x^2 −3x) mvc method give z^′ =K^′ (x+1)e^((3/2)x^2 −3x) +K{ e^((3/2)x^2 −3x) +(x+1)(3x−3) e^((3/2)x^2 −3x) } ={ K^′ (x+1) + K +3K(x+1)(x−1) } e^((3/2)x^2 −3x) (e) ⇒{(x+1)^2 K^′ +K(x+1) +3K(x+1)^2 (x−1)} e^((3/2)x^2 −3x) (2−3x^2 )K(x+1)e^((3/2)x^2 −3x) =xsinx ⇒ (x+1)K^′ +K +3K(x^2 −1) +K(2−3x^2 ) =((xsinx)/(x+1)) ⇒ (x+1)K^′ +K −K =((xsinx)/(x+1)) ⇒K^′ =((xsinx)/((x+1)^2 )) ⇒ K(x) =∫ ((xsinx)/((x+1)^2 ))dx +c_0 ⇒z(x) =(x+1)e^((3/2)x^2 −3x) { ∫ ((xsinx)/((x+1)^2 ))dx +c_0 } =c_0 (x+1) e^((3/2)x^2 −3x) +(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 )) dt we have y^′ (x)=z(x) ⇒y(x) =∫ z(x)dx +λ ⇒ y(x) =∫ c_0 (x+1)e^((3/2)x^2 −3x) dx +∫ {(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 ))dt}dx +λ .](https://www.tinkutara.com/question/Q53965.png)
$${solve}\:\:\left({x}+\mathrm{1}\right){y}^{''} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){y}^{'} \:={xsin}\left({x}\right) \\ $$$${let}\:{y}^{'} ={z}\:\:\:{so}\:\left({e}\right)\:\Leftrightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){z}\:={xsinx}\left({e}\right)\:{let}\:{first}\:{find}\:{z} \\ $$$$\left({he}\right)\:\rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){z}\:=\mathrm{0}\:\Rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:=\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right){z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\:\int\:\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{{x}+\mathrm{1}}\:{dx}\:+{c}\:\:=\int\:\frac{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}{{x}+\mathrm{1}}+{c}\:=\int\mathrm{3}\left({x}−\mathrm{1}\right){dx}\:+\int\:\frac{{dx}}{{x}+\mathrm{1}}\:+{c} \\ $$$$\left.=\mathrm{3}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)+{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow{z}\:={K}\mid{x}+\mathrm{1}\mid{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:{solution}\:{on}\right]−\mathrm{1},+\infty\left[\:\Rightarrow\right. \\ $$$${z}\left({x}\right)\:={K}\left({x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:\:{mvc}\:{method}\:{give} \\ $$$${z}^{'} ={K}^{'} \left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+{K}\left\{\:\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+\left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{3}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \right\} \\ $$$$=\left\{\:{K}^{'} \left({x}+\mathrm{1}\right)\:+\:{K}\:+\mathrm{3}{K}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\:\right\}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \\ $$$$\left({e}\right)\:\Rightarrow\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} {K}^{'} \:+{K}\left({x}+\mathrm{1}\right)\:+\mathrm{3}{K}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{1}\right)\right\}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \\ $$$$\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){K}\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:={xsinx}\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){K}^{'} \:+{K}\:+\mathrm{3}{K}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:+{K}\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right)\:=\frac{{xsinx}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){K}^{'} \:+{K}\:−{K}\:=\frac{{xsinx}}{{x}+\mathrm{1}}\:\Rightarrow{K}^{'} \:=\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int\:\:\:\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:+{c}_{\mathrm{0}} \:\Rightarrow{z}\left({x}\right)\:=\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \left\{\:\int\:\:\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:+{c}_{\mathrm{0}} \right\} \\ $$$$={c}_{\mathrm{0}} \left({x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\int_{.} ^{{x}} \:\:\:\frac{{tsint}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\ $$$${we}\:{have}\:{y}^{'} \left({x}\right)={z}\left({x}\right)\:\Rightarrow{y}\left({x}\right)\:=\int\:{z}\left({x}\right){dx}\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\int\:{c}_{\mathrm{0}} \left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} {dx}\:+\int\:\:\left\{\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\int_{.} ^{{x}} \:\:\frac{{tsint}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right\}{dx}\:+\lambda\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
![(dy/dx)=p (x+1)(dp/dx)+(2−3x^2 )p=xsinx (dp/dx)+((2−3x^2 )/(x+1))p=((xsinx)/(x+1)) I.F=e^(∫((2−3x^2 )/(x+1))dx) =e^(∫[(−3x+3)+((−1)/(x+1))]dx) e^(((−3x^2 )/2)+3x−ln(x+1)) e^(((−3x^2 )/2)+3x) ×e^(−ln(x+1)) e^(((−3x^2 )/2)+3x) ×(1/(x+1)) e^(((−3x^2 )/2)+3x) ×(1/(x+1))(dp/dx)+((2−3x^2 )/(x+1))p×(1/(x+1))×e^(((−3x^2 )/2)+3x) =((xsinx)/(x+1))×e^(−((3x^2 )/2)+3x) ×(1/(x+1)) (d/dx)(p×(e^(((−3x^2 )/2)+3x) /(x+1)))=((xsinx×e^(((−3x^2 )/2)+3x) )/((x+1)^2 )) ∫d(p×(e^(−((3x^2 )/2)+3x) /(x+1)))=∫((xsinx×e^(((−3x^2 )/2)+3x) )/((x+1)^2 ))dx LHS=∫d(p×(e^(−((3x^2 )/2)+3x) /(x+1)))=p×(e^(−((3x^2 )/2)+3x) /(x+1)) Wait pls...](https://www.tinkutara.com/question/Q54027.png)
$$\frac{{dy}}{{dx}}={p} \\ $$$$\left({x}+\mathrm{1}\right)\frac{{dp}}{{dx}}+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){p}={xsinx} \\ $$$$\frac{{dp}}{{dx}}+\frac{\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} }{{x}+\mathrm{1}}{p}=\frac{{xsinx}}{{x}+\mathrm{1}} \\ $$$${I}.{F}={e}^{\int\frac{\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} }{{x}+\mathrm{1}}{dx}} ={e}^{\int\left[\left(−\mathrm{3}{x}+\mathrm{3}\right)+\frac{−\mathrm{1}}{{x}+\mathrm{1}}\right]{dx}} \\ $$$${e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}−{ln}\left({x}+\mathrm{1}\right)} \\ $$$${e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} ×{e}^{−{ln}\left({x}+\mathrm{1}\right)} \\ $$$${e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} ×\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} ×\frac{\mathrm{1}}{{x}+\mathrm{1}}\frac{{dp}}{{dx}}+\frac{\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} }{{x}+\mathrm{1}}{p}×\frac{\mathrm{1}}{{x}+\mathrm{1}}×{e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} =\frac{{xsinx}}{{x}+\mathrm{1}}×{e}^{−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} ×\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\frac{{d}}{{dx}}\left({p}×\frac{{e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} }{{x}+\mathrm{1}}\right)=\frac{{xsinx}×{e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int{d}\left({p}×\frac{{e}^{−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} }{{x}+\mathrm{1}}\right)=\int\frac{{xsinx}×{e}^{\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${LHS}=\int{d}\left({p}×\frac{{e}^{−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} }{{x}+\mathrm{1}}\right)={p}×\frac{{e}^{−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{3}{x}} }{{x}+\mathrm{1}} \\ $$$$\boldsymbol{{W}}{ait}\:{pls}… \\ $$$$ \\ $$