Question Number 183638 by ali009 last updated on 27/Dec/22 $${find}\:{y}\left({t}\right) \\ $$$${y}\left({t}\right)+\mathrm{2}{e}^{{t}} \int_{\mathrm{0}} ^{{t}} \left({y}\left(\tau\right)−{e}^{−\tau} \right){d}\tau={te}^{{t}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 183533 by MikeH last updated on 26/Dec/22 Commented by MikeH last updated on 26/Dec/22 $$\mathrm{please}\:\mathrm{guys}\:\mathrm{solve}\:\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{undetermined} \\ $$$$\mathrm{coefficients}\:\mathrm{or}\:\mathrm{variation}\:\mathrm{of}\:\mathrm{parameters}. \\ $$ Answered by qaz last…
Question Number 183532 by MikeH last updated on 26/Dec/22 Answered by CElcedricjunior last updated on 26/Dec/22 $$\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{{y}}=\boldsymbol{{lnx}}+\mathrm{1}\:\:\:\:\boldsymbol{{y}}\left(\mathrm{1}\right)=−\mathrm{2} \\ $$$$\left(\boldsymbol{{h}}\right):\mathrm{3}\boldsymbol{{xy}}'−\boldsymbol{{y}}=\mathrm{0}=>\boldsymbol{{y}}=\boldsymbol{{e}}^{\int\frac{\mathrm{1}}{\mathrm{3}\boldsymbol{{x}}}\boldsymbol{{dx}}} =\boldsymbol{{ke}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\mid\boldsymbol{{x}}\mid} =\boldsymbol{{ke}}^{\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{lnx}}} \boldsymbol{{pour}}\:\boldsymbol{{x}}\in\mathbb{R}_{+} ^{\ast} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{k}}\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}}…
Question Number 117759 by Dwaipayan Shikari last updated on 13/Oct/20 $$\frac{{d}\theta}{{dx}}=\frac{\sqrt{\mathrm{1}−\frac{\theta^{\mathrm{2}} }{{x}^{\mathrm{2}} }}}{{sin}\left(\theta+{x}\right)} \\ $$ Commented by TANMAY PANACEA last updated on 13/Oct/20 $${assuming}\:\theta\:{and}\:{x}\:{in}\:{radian} \\…
Question Number 52208 by SUJIT420 last updated on 04/Jan/19 Commented by maxmathsup by imad last updated on 04/Jan/19 $${its}\:{not}\:{c}\:{its}\:{e}\:. \\ $$ Commented by maxmathsup by…
Question Number 117728 by syamil last updated on 13/Oct/20 $${Solution}\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{3}\frac{{dy}}{{dx}}\:−\:\mathrm{4}{y}\:=\:{x}^{\mathrm{2}} \\ $$ Answered by TANMAY PANACEA last updated on 13/Oct/20 $${y}={e}^{{mx}} \\ $$$$\left({m}^{\mathrm{2}}…
Question Number 117632 by syamil last updated on 12/Oct/20 $${Solution}\:{from}\:\:\:\mathrm{2}{xy}\:{dy}\:=\:\left({x}^{\mathrm{2}\:} \:−\:{y}^{\mathrm{2}} \right){dx} \\ $$ Answered by 1549442205PVT last updated on 13/Oct/20 $$\:\mathrm{2}{xy}\:{dy}\:=\:\left({x}^{\mathrm{2}\:} \:−\:{y}^{\mathrm{2}} \right){dx} \\…
Question Number 117608 by TANMAY PANACEA last updated on 12/Oct/20 $${solve} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{w}^{\mathrm{2}} {x}=\mathrm{0} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 117602 by TANMAY PANACEA last updated on 12/Oct/20 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{a}^{\mathrm{2}} {y}={cosax} \\ $$ Answered by TANMAY PANACEA last updated on 12/Oct/20 $${y}={e}^{{mx}}…
Question Number 117585 by syamil last updated on 12/Oct/20 $${solution}\:\:\:\:\frac{{dy}}{{dx}}\:=\:{sin}\:{x}\:+\:{e}^{\mathrm{2}{x}} \:+\:{x}^{\mathrm{2}} \\ $$ Commented by TANMAY PANACEA last updated on 12/Oct/20 $${i}\:{can}\:{not}\:{post}\:{question} \\ $$$${so}\:{here}\:{i}\:{am}\:{posting}?{question} \\…