Question Number 157267 by cortano last updated on 21/Oct/21 $$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}={x}\: \\ $$ Commented by MathsFan last updated on 22/Oct/21 $${any}\:{caption}???? \\ $$ Commented by…
Question Number 26191 by sorour87 last updated on 21/Dec/17 $$\mathrm{2}\frac{{dy}}{{dx}}+{x}=\mathrm{4}\sqrt{{y}} \\ $$ Answered by mrW1 last updated on 22/Dec/17 $${y}={u}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{u}\frac{{du}}{{dx}} \\ $$$$\Rightarrow\mathrm{4}{u}\frac{{du}}{{dx}}+{x}=\mathrm{4}{u} \\…
Question Number 91678 by john santu last updated on 02/May/20 $$\left({y}+\mathrm{2}{px}\right)^{\mathrm{2}} \:=\:\mathrm{2}{px}^{\mathrm{2}} \\ $$$${solve}\:{by}\:{Clairaut}'{s}\:{method}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 26127 by sorour87 last updated on 20/Dec/17 $${y}+\mathrm{2}{y}^{\mathrm{3}} {y}^{\left(\mathrm{1}\right)} =\left({x}+\mathrm{4}{y}\mathrm{ln}\:\left({y}\right)\right){y}^{\left(\mathrm{1}\right)} \\ $$ Answered by prakash jain last updated on 21/Dec/17 $${y}+\mathrm{2}{y}^{\mathrm{3}} \frac{{dy}}{{dx}}=\left({x}+\mathrm{4}{y}\mathrm{ln}\:{y}\right)\frac{{dy}}{{dx}} \\…
Question Number 91656 by jagoll last updated on 02/May/20 Commented by john santu last updated on 02/May/20 $$\frac{{dy}}{{dx}}\:=\:\frac{−\mathrm{3}{xy}−{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{xy}}\:=\:\frac{−\frac{\mathrm{3}{y}}{{x}}−\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{y}}{{x}}\right)} \\ $$$$\left[\:{y}\:=\:{px}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{p}\:+\:{x}\frac{{dp}}{{dx}}\:\right] \\ $$$${p}\:+{x}\:\frac{{dp}}{{dx}}\:=\:\frac{−\mathrm{3}{p}−{p}^{\mathrm{2}}…
Question Number 91643 by jagoll last updated on 02/May/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{y}−{x}}{{x}}\: \\ $$$$ \\ $$ Commented by john santu last updated on 02/May/20 Commented by Prithwish…
Question Number 157141 by cortano last updated on 20/Oct/21 $$\:\begin{cases}{\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{2}\:\frac{{dy}}{{dx}}+{y}\:=\mathrm{3}{e}^{\mathrm{4}{x}} }\\{{y}\left(\mathrm{0}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\:;\:\frac{{dy}}{{dx}}\mid_{{x}=\mathrm{0}} =\:\frac{\mathrm{13}}{\mathrm{3}}}\end{cases}\: \\ $$ Commented by john_santu last updated on 21/Oct/21 $$\:\begin{cases}{{y}''−\mathrm{2}{y}'+{y}=\mathrm{3}{e}^{\mathrm{4}{x}} }\\{{y}\left(\mathrm{0}\right)=−\frac{\mathrm{2}}{\mathrm{3}}\:;\:{y}'\left(\mathrm{0}\right)=\frac{\mathrm{13}}{\mathrm{3}}}\end{cases}…
Question Number 91558 by jagoll last updated on 01/May/20 $$\left({x}^{\mathrm{2}} +\mathrm{1}\right){y}'+{y}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0}\: \\ $$ Commented by jagoll last updated on 01/May/20 $${it}\:{Bernoulli}\:{diff}\:{eq}? \\ $$ Answered…
Question Number 25981 by ifcrna380w last updated on 17/Dec/17 Answered by prakash jain last updated on 17/Dec/17 $${a}^{\mathrm{2}} \left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} ={y} \\ $$$$\frac{{dy}}{{dx}}=\pm\frac{\sqrt{{y}}}{{a}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\sqrt{{y}}}{{a}} \\…
Question Number 157033 by amin96 last updated on 18/Oct/21 $${y}''=−{y} \\ $$ Commented by aliyn last updated on 19/Oct/21 $$\boldsymbol{{y}}^{''} \:+\boldsymbol{{y}}\:=\:\mathrm{0}\:\Rightarrow\:\left(\:\boldsymbol{{m}}−\boldsymbol{{i}}\right)\:\left(\boldsymbol{{m}}+\boldsymbol{{i}}\right)\:=\:\mathrm{0}\:\Rightarrow\:\boldsymbol{{m}}=\pm\boldsymbol{{i}} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{y}}\:=\:\boldsymbol{{c}}_{\mathrm{1}}…