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Question Number 108384 by bobhans last updated on 16/Aug/20

$$\left(1\right)\int \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx?$$$$\left(2\right)\frac{d^{2}y}{{dx}^2}-2\frac{dy}{dx}+y=e^x$$

Answered by bemath last updated on 16/Aug/20

Commented by bobhans last updated on 16/Aug/20

$$yes.....$$

Answered by Sarah85 last updated on 16/Aug/20

$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$$$$\mathrm{let}t=\sqrt{\tan x}\iff x=\arctan t^2\iff dx=\frac{2t}{t^4+1}dt$$$$\int \frac{2t^2}{(t+1)(t^4+1)}dt=$$$$=\int \frac{1}{t+1}dt-\int \frac{t^3-t^2-t+1}{t^4+1}dt$$$$=\int \frac{1}{t+1}dt-\int \frac{(1-\sqrt{2})(t+1)}{2(t^2-\sqrt{2}t+1)}dt-\int \frac{(1+\sqrt{2})(t+1)}{2(t^2+\sqrt{2}t+1)}dt$$$$\mathrm{now}\mathrm{I}\mathrm{use}\mathrm{the}\mathrm{obvious}\mathrm{formulas}\mathrm{to}\mathrm{get}$$$$\ln \mid t+1\mid -\frac{1-\sqrt{2}}{4}\ln (t^2-\sqrt{2}t+1)+\frac{1}{2}\arctan (\sqrt{2}t-1)-\frac{1+\sqrt{2}}{4}\ln (t^2+\sqrt{2}t+1)-\frac{1}{2}\arctan (\sqrt{2}t+1)$$$$\mathrm{now}\mathrm{insert}t=\sqrt{\tan x}$$

Answered by mathmax by abdo last updated on 17/Aug/20

$$2){\mathrm{y}}^{{}^{″}}-{2\mathrm{y}}^{{}^{\prime }}+\mathrm{y}={\mathrm{e}}^{\mathrm{x}}$$$$\mathrm{h}\to {\mathrm{r}}^2-2\mathrm{r}+1=0\Rightarrow {(\mathrm{r}-1)}^2=0\Rightarrow \mathrm{r}=1\left(\mathrm{double}\right)\Rightarrow {\mathrm{y}}_{\mathrm{h}}=(\mathrm{ax}+\mathrm{b}){\mathrm{e}}^{\mathrm{x}}$$$$={\mathrm{axe}}^{\mathrm{x}}+{\mathrm{be}}^{\mathrm{x}}={\mathrm{au}}_1+{\mathrm{bu}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}{\mathrm{xe}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\\ (\mathrm{x}+1){\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\end{array}\right|={\mathrm{xe}}^{2\mathrm{x}}-(\mathrm{x}+1){\mathrm{e}}^{2\mathrm{x}}=-{\mathrm{e}}^{2\mathrm{x}}\ne 0$$$${\mathrm{W}}_1=\left|\begin{array}{c}\mathrm{o}{\mathrm{e}}^{\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\end{array}\right|=-{\mathrm{e}}^{2\mathrm{x}}$$$${\mathrm{W}}_2=\left|\begin{array}{c}{\mathrm{xe}}^{\mathrm{x}}0\\ (\mathrm{x}+1){\mathrm{e}}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}}\end{array}\right|={\mathrm{xe}}^{2\mathrm{x}}$$$${\mathrm{V}}_1=\int \frac{{\mathrm{W}}_1}{\mathrm{W}}\mathrm{dx}=\int \frac{-{\mathrm{e}}^{2\mathrm{x}}}{-{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}=\mathrm{x}$$$${\mathrm{V}}_2=\int \frac{{\mathrm{W}}_2}{\mathrm{W}}\mathrm{dx}=\int \frac{{\mathrm{xe}}^{2\mathrm{x}}}{-{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}=-\frac{{\mathrm{x}}^2}{2}\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2={\mathrm{xe}}^{\mathrm{x}}\left(\mathrm{x}\right)+{\mathrm{e}}^{\mathrm{x}}(-\frac{{\mathrm{x}}^2}{2})=\frac{{\mathrm{x}}^2}{2}{\mathrm{e}}^{\mathrm{x}}\Rightarrow \mathrm{the}\mathrm{general}\mathrm{solution}$$$$\mathrm{is}\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}=(\mathrm{ax}+\mathrm{b}){\mathrm{e}}^{\mathrm{x}}+\frac{{\mathrm{x}}^2}{2}{\mathrm{e}}^{\mathrm{x}}$$

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