Question Number 74142 by MASANJAJ last updated on 19/Nov/19 Answered by Rio Michael last updated on 19/Nov/19 $$\:{T}_{\mathrm{7}} \:=\:{a}\:+\:\mathrm{6}{d}\:=\:\mathrm{6}\:−−−\left(\mathrm{1}\right) \\ $$$${T}_{\mathrm{18}} \:=\:{a}\:+\:\mathrm{17}{d}\:=\:\mathrm{22}−−−\left(\mathrm{2}\right) \\ $$$$\:{eqn}\left(\mathrm{2}\right)\:−\:{eqn}\left(\mathrm{1}\right)\:\Rightarrow\:\mathrm{11}{d}\:=\:\mathrm{16} \\…
Question Number 8515 by Basant007 last updated on 14/Oct/16 $$\mathrm{solve}\:\:\mathrm{y}=\mathrm{px}+\mathrm{p}^{\mathrm{3}} ,\:\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$ Commented by prakash jain last updated on 14/Oct/16 $$\mathrm{differentiaing} \\ $$$${p}={p}+{x}\frac{{dp}}{{dx}}+\mathrm{3}{p}^{\mathrm{2}} \frac{{dp}}{{dx}}…
Question Number 139426 by I want to learn more last updated on 26/Apr/21 Answered by physicstutes last updated on 26/Apr/21 $$\mathrm{1}.\:\:{x}\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{2}} +{y} \\ $$$${y}\:=\:{x}^{\mathrm{2}} +{cx}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}\:+\:{c}…
Question Number 139159 by rs4089 last updated on 23/Apr/21 $${solve}\:{PDE}\:\:\:\:\:\:\:\:\:{py}^{\mathrm{3}} +{qx}^{\mathrm{2}} =\mathrm{0}\:\: \\ $$ Commented by Dwaipayan Shikari last updated on 23/Apr/21 $$\frac{\partial^{\mathrm{3}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{3}} }+\frac{\partial^{\mathrm{2}}…
Question Number 8043 by eva suting last updated on 28/Sep/16 $${solve}\:\left({xz}+{y}^{\mathrm{2}} \right)+\left(\mathrm{yz}−\mathrm{zx}^{\mathrm{2}} \right)\mathrm{q}+\mathrm{2xy}+\mathrm{z}^{\mathrm{2}} =\mathrm{0} \\ $$ Commented by prakash jain last updated on 28/Sep/16 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{which}\:\mathrm{variable}?…
Question Number 73495 by ajfour last updated on 13/Nov/19 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }={c}^{\mathrm{2}} {x}^{\mathrm{2}} {y}\:\:\:\:\:\left({y}={a}\:,\:{x}=\mathrm{0}\:\right) \\ $$ Answered by mind is power last updated on 13/Nov/19…
Question Number 7628 by Tawakalitu. last updated on 06/Sep/16 $${By}\:{the}\:{use}\:{of}\:{substitution}\:\:{x}\:=\:\mu^{\mathrm{2}} ,\:{show}\:{that} \\ $$$${the}\:{legendary}\:{equation}\:, \\ $$$$\left(\mathrm{1}\:−\:\mu^{\mathrm{2}} \right){y}''\:−\:\mathrm{2}\mu{y}'\:+\:{n}\left({n}\:+\:\mathrm{1}\right){y}\:=\:\mathrm{0},\: \\ $$$${where}\:{n}\:{is}\:{a}\:{constant}\:{change}\:{to}\:{hyper}\:{geometric} \\ $$$${equation}\:.\:{hence}\:{obtain}\:{the}\:{solution}\:{to}\:{the}\: \\ $$$${resulting}\:{hyper}\:{geometric}\:{differential}\:{equation}\: \\ $$$${by}\:{way}\:{of}\:{comparison}. \\…
Question Number 73080 by oyemi kemewari last updated on 06/Nov/19 $$\mathrm{y}''=\mathrm{e}^{\mathrm{y}} \\ $$$$\mathrm{pls}\:\mathrm{solve} \\ $$ Answered by mind is power last updated on 06/Nov/19 $$\mathrm{y}''.\mathrm{y}'=\mathrm{y}'\mathrm{e}^{\mathrm{y}}…
Question Number 7216 by peter james last updated on 16/Aug/16 Answered by Yozzia last updated on 16/Aug/16 $${ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\frac{\mathrm{3}−{y}}{{x}+\mathrm{3}}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${Let}\:{u}=\frac{{x}+{y}}{{x}+\mathrm{3}}\Rightarrow{y}={u}\left({x}+\mathrm{3}\right)−{x} \\ $$$$\therefore\:{y}'={u}'\left({x}+\mathrm{3}\right)+{u}−\mathrm{1} \\ $$$$\Rightarrow{y}'+\mathrm{1}={u}+{u}'\left({x}+\mathrm{3}\right). \\…
Question Number 138240 by bobhans last updated on 11/Apr/21 $$\:\left({x}−\mathrm{1}\right)\frac{{dy}}{{dx}}\:+{xy}\:=\:\mathrm{2}{xe}^{−{x}} \\ $$ Answered by EDWIN88 last updated on 11/Apr/21 $$\:\frac{{dy}}{{dx}}\:+\:\frac{{x}}{{x}−\mathrm{1}}\:{y}\:=\:\frac{\mathrm{2}{x}}{{e}^{{x}} \left({x}−\mathrm{1}\right)} \\ $$$${put}\:{IF}\:=\:{e}^{\int\:\frac{{x}}{{x}−\mathrm{1}}\:{dx}} \:=\:{e}^{\int\:\frac{{x}−\mathrm{1}+\mathrm{1}}{{x}−\mathrm{1}}\:{dx}} ={e}^{{x}+\mathrm{ln}\:\left({x}−\mathrm{1}\right)}…