Question Number 324 by 123456 last updated on 25/Jan/15 $${u}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R} \\ $$$$\frac{\partial{u}}{\partial{x}}+\frac{\partial{u}}{\partial{y}}=\frac{\left({x}+{y}\right){u}}{{xy}} \\ $$ Answered by prakash jain last updated on 21/Dec/14 $$\mathrm{The}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{symmetric}\:\mathrm{in}\:{x},{y}. \\…
Question Number 304 by 123456 last updated on 20/Dec/14 $${u}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R} \\ $$$$\frac{\partial{u}}{\partial{y}}+\frac{\partial{u}}{\partial{x}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{xu}=\mathrm{0} \\ $$ Commented by prakash jain last updated on 20/Dec/14…
Question Number 297 by 123456 last updated on 25/Jan/15 $${u}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${v}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\begin{cases}{{u}=\frac{{dv}}{{dx}}}\\{{v}=\frac{{du}}{{dx}}}\end{cases} \\ $$ Answered by prakash jain last updated on 19/Dec/14 $${u}=\frac{{dv}}{{dx}}=\frac{{d}^{\mathrm{2}}…
Question Number 293 by 123456 last updated on 25/Jan/15 $${u}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R} \\ $$$$\frac{\partial{u}}{\partial{x}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{xu}=\mathrm{0} \\ $$ Answered by prakash jain last updated on 19/Dec/14…
Question Number 289 by 123456 last updated on 25/Jan/15 $${x}^{\mathrm{3}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}^{\mathrm{2}} \frac{{dy}}{{dx}}+{xy}=\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${y}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${y}'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{5} \\ $$ Answered by prakash jain…
Question Number 271 by 123456 last updated on 25/Jan/15 $$\:{x}^{\mathrm{3}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{x}^{\mathrm{2}} \frac{{dy}}{{dx}}+{xy}=\mathrm{2}−{x} \\ $$$${y}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${y}'\left(\mathrm{1}\right)=−\mathrm{1} \\ $$ Answered by prakash jain last…
Question Number 250 by 123456 last updated on 25/Jan/15 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{2}{x}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$ Answered by prakash jain last updated on 17/Dec/14 $$\frac{{dy}}{{dx}}={u} \\…
Question Number 131315 by Engr_Jidda last updated on 03/Feb/21 $${Given}\:{that}\:{y}_{{n}} \left({t}\right)=\varrho^{{t}} {B}_{{n}} {sin}\frac{{n}\pi}{\mathrm{4}}{t} \\ $$$${Evaluate}\:\int_{\mathrm{0}} ^{\mathrm{4}} \left[{y}_{{n}} \left({t}\right)\right]^{\mathrm{2}} {dt} \\ $$ Answered by Ar Brandon…
Question Number 236 by 123456 last updated on 25/Jan/15 $$\mathrm{solve} \\ $$$${y}''−{y}'−{y}={e}^{{x}} +{e}^{\mathrm{2}{x}} \\ $$$$\begin{cases}{{y}\left(\mathrm{0}\right)+{y}'\left(\mathrm{0}\right)=\mathrm{1}}\\{{y}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)=\mathrm{0}}\end{cases} \\ $$ Answered by prakash jain last updated on 17/Dec/14…
Question Number 224 by 123456 last updated on 25/Jan/15 $$\mathrm{solve} \\ $$$${x}\frac{{dy}}{{dx}}+{y}=\alpha{x}+\beta \\ $$ Answered by mreddy last updated on 16/Dec/14 $$\frac{{dy}}{{dx}}+\frac{{y}}{{x}}=\frac{\alpha{x}+\beta}{{x}} \\ $$$$\mathrm{Integrating}\:\mathrm{Factor}={e}^{\int\frac{\mathrm{1}}{{x}}{dx}} ={e}^{\mathrm{ln}\:{x}}…