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}} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{y}}\:=\:\boldsymbol{{c}}_{\mathrm{1}}…
Question Number 91422 by jagoll last updated on 30/Apr/20 $$\frac{{dy}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 91417 by jagoll last updated on 30/Apr/20 $${y}'+\mathrm{2}{y}={e}^{−{x}} \\ $$ Commented by john santu last updated on 30/Apr/20 $${IF}\:{u}\left({x}\right)=\:{e}^{\int\:\mathrm{2}\:{dx}} \:=\:{e}^{\mathrm{2}{x}} \\ $$$${solution}\: \\…
Question Number 156931 by cortano last updated on 17/Oct/21 $$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}''−\mathrm{4}{xy}'−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}=\mathrm{0} \\ $$ Commented by john_santu last updated on 17/Oct/21 $${y}=\frac{\mathrm{sin}\:{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{c} \\ $$…
Question Number 91384 by jagoll last updated on 30/Apr/20 $${particular}\:{integral}\: \\ $$$${y}''+\mathrm{3}{y}'+\mathrm{2}{y}\:=\:\mathrm{4cos}\:^{\mathrm{2}} {x} \\ $$ Commented by jagoll last updated on 30/Apr/20 $${PI}\:\Rightarrow{y}_{{p}} \:=\:\frac{\mathrm{4cos}\:^{\mathrm{2}} {x}}{{D}^{\mathrm{2}}…