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Question Number 71516 by oyemi kemewari last updated on 16/Oct/19

$${5\mathrm{y}}^{″}={(1+{\mathrm{y}}^{\prime 2})}^{\frac{3}{2}}$$$$\mathrm{please}\mathrm{solve}\mathrm{the}\mathrm{differtial}\mathrm{equation}$$

Answered by mind is power last updated on 17/Oct/19

$$\mathrm{let}\mathrm{z}={\mathrm{y}}^{\prime }$$$${5\mathrm{z}}^{\prime }={(1+{\mathrm{z}}^2)}^{\frac{3}{2}}$$$$\Rightarrow \frac{\mathrm{dz}}{{(1+{\mathrm{z}}^2)}^{\frac{3}{2}}}=\mathrm{dx}$$$$\mathrm{z}=\mathrm{tg}\left(\mathrm{u}\right)$$$$\mathrm{dz}=(1+{\mathrm{tg}}^2\left(\mathrm{u}\right))\mathrm{du}$$$$\Rightarrow \frac{\mathrm{du}}{\sqrt{1+{\mathrm{tg}}^2\left(\mathrm{u}\right)}}=\frac{1}{5}\mathrm{dx}$$$$\Rightarrow \cos \left(\mathrm{u}\right)=\frac{\mathrm{x}}{5}+\mathrm{v}$$$$\mathrm{u}=\mathrm{arcos}(\frac{\mathrm{x}}{5}+\mathrm{v})$$$$\mathrm{z}=\mathrm{tg}\left(\mathrm{u}\right)=\frac{\sqrt{1-{(\frac{\mathrm{x}}{5}+\mathrm{v})}^2}}{\frac{\mathrm{x}}{5}+\mathrm{v}}$$$$\mathrm{y}=\int \mathrm{zdx}=\int \frac{\sqrt{1-(\frac{\mathrm{x}}{5}+\mathrm{v})}}{\frac{\mathrm{x}}{5}+\mathrm{v}}\mathrm{dx}$$$$=5\int \frac{\sqrt{1-\mathrm{u}}}{\mathrm{u}}\mathrm{du},\sqrt{1-\mathrm{u}}=\mathrm{r}\Rightarrow \mathrm{du}=-2\mathrm{rdr}$$$$5\int \frac{-{2\mathrm{r}}^2}{1-{\mathrm{r}}^2}\mathrm{dr}=-10\int -1+\frac{1}{1-{\mathrm{r}}^2}\mathrm{dr}=10\mathrm{r}-10\int (\frac{1}{2(1-\mathrm{r})}+\frac{1}{2(1+\mathrm{r})})\mathrm{dr}$$$$=10\mathrm{r}-5\ln \mid 1-{\mathrm{r}}^2\mid$$$$\mathrm{y}=10\sqrt{1-(\frac{\mathrm{x}}{5}+\mathrm{v})}-5\ln \mid \frac{\mathrm{x}}{5}+\mathrm{v}\mid +\mathrm{c}$$$$$$$$$$

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