Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 73488 by abdomathmax last updated on 13/Nov/19

$$solve{xy}^{{}^{″}}+(x^2-1)y^{{}^{\prime }}=x{e}^{-x^2}$$

Commented by mathmax by abdo last updated on 14/Nov/19

$$let{y}^{{}^{\prime }}=z\left(e\right)\Rightarrow {xz}^{{}^{\prime }}+(x^2-1)z={xe}^{-x^2}$$$$\left(he\right)\to {xz}^{{}^{\prime }}+(x^2-1)z=o\Rightarrow {xz}^{{}^{\prime }}=(1-x^2)z\Rightarrow \frac{z^{{}^{\prime }}}{z}=\frac{1-x^2}{x}\Rightarrow$$$$\frac{z^{{}^{\prime }}}{z}=\frac{1}{x}-x\Rightarrow ln\mid z\mid =ln\mid x\mid -\frac{x^2}{2}+c\Rightarrow z=k\mid x\mid e^{-\frac{x^2}{2}}(wesearchsolutin$$$$on[0,+\mathrm{\infty }\left[\right)\Rightarrow z=kx{e}^{-\frac{x^2}{2}}$$$$letusemvcmethodwehave{z}^{{}^{\prime }}=k^{{}^{\prime }}{xe}^{-\frac{x^2}{2}}$$$$+k\{e^{-\frac{x^2}{2}}+x(-x)e^{-\frac{x^2}{2}}\}=k^{{}^{\prime }}x{e}^{-\frac{x^2}{2}}+k(1-x^2)e^{-\frac{x^2}{2}}$$$$\left(e\right)\Rightarrow k^{{}^{\prime }}{x}^2{e}^{-\frac{x^2}{2}}+k(1-x^2)e^{-\frac{x^2}{2}}+(x^2-1){kxe}^{-\frac{x^2}{2}}={xe}^{-x^2}\Rightarrow$$$$k^{{}^{\prime }}=\frac{{xe}^{-x^2}}{x^2{e}^{-\frac{x^2}{2}}}=\frac{1}{x}{e}^{-x^2+\frac{x^2}{2}}=\frac{1}{x}{e}^{-\frac{x^2}{2}}\Rightarrow$$$$k\left(x\right)={\int }_1^x\frac{e^{-\frac{t^2}{2}}}{t}dt+c(c=k\left(1\right)wehavez\left(1\right)=k\left(1\right)e^{-\frac{1}{2}}\Rightarrow$$$$k\left(1\right)=z\left(1\right)e^{\frac{1}{2}}wehavez\left(x\right)=xk\left(x\right)e^{-\frac{x^2}{2}}$$$$={xe}^{-\frac{x^2}{2}}\{{\int }_1^x\frac{e^{-\frac{t^2}{2}}}{t}dt+z\left(1\right)e^{\frac{1}{2}}\}wehave{y}^{{}^{\prime }}=z\Rightarrow$$$$y\left(x\right)={\int }_{.}^{x}z\left(u\right)du+C$$$$={\int }_{.}^x\left\{u{e}^{-\frac{u^2}{2}}\{{\int }_1^u\frac{e^{-\frac{t^2}{2}}}{t}dt+z\left(1\right)\sqrt{e}\}\right\}du+C$$

Answered by mind is power last updated on 13/Nov/19

$$y^{\prime }=z$$$$homgen$$$${xz}^{\prime }+(x^2-1)z=0$$$$\Rightarrow \frac{z^{\prime }}{z}=\frac{1}{x}-x\Rightarrow ln\mid z\mid =-\frac{x^2}{2}+ln\left(x\right)\Rightarrow z={kxe}^{-\frac{x^2}{2}}$$$${xz}^{\prime }+(x^2-1)z={xe}^{-x^2}$$$$x(k^{\prime }{xe}^{-\frac{x^2}{2}}+(e^{-\frac{x^2}{2}}-x^2{e}^{\frac{-x^2}{2}})k)+(x^2-1){kxe}^{-\frac{x^2}{2}}={xe}^{-x^2}$$$$\Rightarrow k^{\prime }{x}^2{e}^{-\frac{x^2}{2}}={xe}^{-x^2}$$$$k^{\prime }=\frac{e^{-\frac{x^2}{2}}}{x}$$$$k=\int \frac{e^{-x^2}}{x}dx$$$$x=\sqrt{t}\Rightarrow dx=\frac{dt}{2\sqrt{t}}$$$$\int \frac{e^{-t}}{2t}dt=\int \frac{e^t}{2t}dt=\frac{1}{2}{E}_i(-t)$$$$E_i=exponentialintegralfunction$$$$\Rightarrow k=\frac{Ei(-x^2)}{2}+c$$$$\Rightarrow z\left(x\right)={xe}^{-\frac{x^2}{2}}\{\int \frac{e^{-x^2}}{x}dx+c\}=\frac{1}{2}{xe}^{-\frac{x^2}{2}}Ei(-x^2)+c\frac{{xe}^{-\frac{x^2}{2}}}{2}$$$$y=\int z\left(x\right)dx$$$$bypart=-\frac{e^{-\frac{x^2}{2}}}{2}Ei(-x^2)+\frac{1}{2}\int e^{-\frac{x^2}{2}}.\frac{e^{-x^2}}{x}dx-\frac{c}{4}{xe}^{-\frac{x^2}{2}}$$$$=-\frac{e^{-\frac{x^2}{2}}}{2}{E}_i(-x^2)+\frac{1}{2}\int \frac{e^{-\left(\frac{3}{2}\right)x^2}}{x}dx$$$$\frac{3}{2}{x}^2=t\Rightarrow \frac{1}{2}\int \frac{e^{-t}}{t}.dt=\frac{1}{2}{E}_i(-t)=\frac{1}{2}{E}_i(-\frac{3}{2}{x}^2)+d$$$$y\left(x\right)=-\frac{e^{-\frac{x^2}{2}}}{2}{E}_i(-x^2)+\frac{1}{4}{E}_i(-\frac{3x^2}{2})-\frac{{cxe}^{-\frac{x^2}{2}}}{4}+d$$$$c,dconstante$$$$$$$$$$$$$$

Answered by arkanmath7@gmail.com last updated on 13/Nov/19

$$let{y}^{{}^{\prime }}=p,y^{″}=p^{{}^{\prime }}$$$$xp{}^{{}^{\prime }}+(x^2-1)p={xe}^{-x^2}/x$$$$p^{{}^{\prime }}+(x-\frac{1}{x})p=e^{-x^2}$$$$P\left(x\right)=x-\frac{1}{x},Q\left(x\right)=e^{-x^2}$$$$p=e^{-\int P\left(x\right)dx}[\int Q\left(x\right)e^{\int P\left(x\right)dx}dx]$$$$p=e^{-\int (x-\frac{1}{x})dx}[\int e^{-x^2}{e}^{\int (x-\frac{1}{x})dx}dx]$$$$p=e^{-\frac{x^2}{2}+lnx}[\int e^{-x^2}{e}^{\frac{x^2}{2}-lnx}dx]$$$$p=e^{-\frac{x^2}{2}}{e}^{lnx}[\int e^{-x^2}{e}^{\frac{x^2}{2}}{e}^{lnx}dx]$$$$$$$$p={xe}^{-\frac{x^2}{2}}[\int -x{e}^{-\frac{1}{2}{x}^2}dx]$$$$p={xe}^{-\frac{x^2}{2}}[\frac{1}{2}{e}^{-\frac{1}{2}{x}^2}+c]$$$$p=\frac{1}{2}{xe}^{-x^2}+{cxe}^{-\frac{x^2}{2}}$$$$\frac{dy}{dx}=\frac{1}{2}{xe}^{-x^2}+{cxe}^{-\frac{x^2}{2}}$$$$$$$$dy=(\frac{1}{2}{xe}^{-x^2}+{cxe}^{-\frac{x^2}{2}})dx$$$$$$$$\int dy=\int (\frac{1}{2}{xe}^{-x^2}+{cxe}^{-\frac{x^2}{2}})dx$$$$$$$$y=-\frac{1}{4}{e}^{-x^2}-{ce}^{-\frac{x^2}{2}}+c_1$$

Commented by mind is power last updated on 13/Nov/19

$$lin8e^{-ln\left(x\right)}.not{e}^{ln\left(x\right)}$$

Commented by arkanmath7@gmail.com last updated on 13/Nov/19

$$yestrue$$

Commented by arkanmath7@gmail.com last updated on 13/Nov/19

$$mistake!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com