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Question Number 51876 by Tinkutara last updated on 31/Dec/18

Commented by maxmathsup by imad last updated on 31/Dec/18

$$letsolve{y}^{{}^{\prime }}+2y=f\left(x\right)\left(e\right)$$$$\left(he\right)\Rightarrow y^{{}^{\prime }}+2y=0\Rightarrow \frac{y^{{}^{\prime }}}{y}=-2\Rightarrow ln\left(y\right)=-2x+k\Rightarrow y\left(x\right)=C{e}^{-2x}$$$$mvcmethodgive{y}^{{}^{\prime }}=C^{{}^{\prime }}{e}^{-2x}-2C{e}^{-2x}$$$$\left(e\right)\iff C^{{}^{\prime }}{e}^{-2x}-2C{e}^{-2x}+2C{e}^{-2x}=f\left(x\right)\Rightarrow C^{{}^{\prime }}=f\left(x\right)e^{2x}\Rightarrow$$$$C\left(x\right)={\int }_0^{x}f\left(t\right)e^{2t}dt+\lambda with\lambda =C\left(0\right)=y\left(0\right)\Rightarrow$$$$y\left(x\right)=({\int }_0^{x}f\left(t\right)e^{2t}dt+y\left(0\right))e^{-2x}soify\left(o\right)=0\Rightarrow$$$$y\left(x\right)=e^{-2x}{\int }_0^{x}f\left(t\right)e^{2t}dt\Rightarrow y\left(\frac{3}{2}\right)=e^{-3}{\int }_0^{\frac{3}{2}}f\left(t\right)e^{2t}dt$$$$=e^{-3}{\int }_0^1{e}^{2t}dt=\frac{e^{-3}}{2}{\left[e^{2t}\right]}_0^1=\frac{e^{-3}}{2}(e^2-1)=\frac{e^{-1}-e^{-3}}{2}=\frac{e^2-1}{2e^3}.$$$$$$

Commented by Tinkutara last updated on 31/Dec/18

Thanks very much Sir!

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18

$$\frac{dy}{dx}+2y=f\left(x\right)[\frac{dy}{dx}+py=QI.F=e^{\int pdx}]$$$$intregatingfactor{e}^{\int pdx}=e^{\int 2dx}=e^{2x}$$$$e^{2x}\frac{dy}{dx}+e^{2x}\times 2y=e^{2x}f\left(x\right)$$$$e^{2x}\times \frac{d}{dx}\left(y\right)+y\times \frac{d}{dx}\left(e^{2x}\right)=e^{2x}f\left(x\right)$$$$\frac{d}{dx}[y.e^{2x}]=e^{2x}f\left(x\right)$$$$\int d\left[{ye}^{2x}\right]=\int e^{2x}f\left(x\right)dx$$$$$$$${\int }_0^{\frac{3}{2}}d\left[{ye}^{2x}\right]={\int }_0^1{e}^{2x}f\left(x\right)dx+{\int }_1^{\frac{3}{2}}{e}^{2x}f\left(x\right)dx$$$$\mid {ye}^{2x}{\mid }_0^{\frac{3}{2}}={\int }_0^1{e}^{2x}\times 1dx+{\int }_1^{\frac{3}{2}}{e}^{2x}\times 0dx$$$$y\left(\frac{3}{2}\right)e^3-y\left(0\right)e^{2\times 0}=\mid \frac{e^{2x}}{2}{\mid }_0^1$$$$y\left(\frac{3}{2}\right)e^3-0\times e^0=\frac{e^2-1}{2}$$$$y\left(\frac{3}{2}\right)e^3=e^2-1$$$$y\left(\frac{3}{2}\right)=\frac{e^2-1}{2e^3}$$$$$$$$$$$$\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}....\mathit{y}\left(\frac{3}{2}\right)=\frac{{\mathit{e}}^2-1}{2{\mathit{e}}^3}$$$$$$$$$$$$$$$$$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18

$$\mathit{i}\mathit{h}\mathit{a}\mathit{v}\mathit{e}\mathit{r}\mathit{e}\mathit{c}\mathit{t}\mathit{i}\mathit{f}\mathit{i}\mathit{e}\mathit{d}...\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}\mathit{n}\mathit{o}\mathit{w}...\mathit{t}\mathit{h}\mathit{s}\mathit{n}\mathit{k}\mathit{s}$$$$\mathit{f}\mathit{o}\mathit{r}\mathit{r}\mathit{e}\mathit{a}\mathit{d}\mathit{b}\mathit{e}\mathit{t}\mathit{w}\mathit{e}\mathit{e}\mathit{n}\mathit{l}\mathit{i}\mathit{n}\mathit{e}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 31/Dec/18

$$yesitisamistake...letmerectify...$$

Commented by maxmathsup by imad last updated on 31/Dec/18

$$sirTanmayyouransweriscorrect.$$

Commented by Tinkutara last updated on 01/Jan/19

Thank you very much Sir! I got the answer. ��������

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