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)} \\ $$$${put}\:{IF}\:=\:{e}^{\int\:\frac{{x}}{{x}−\mathrm{1}}\:{dx}} \:=\:{e}^{\int\:\frac{{x}−\mathrm{1}+\mathrm{1}}{{x}−\mathrm{1}}\:{dx}} ={e}^{{x}+\mathrm{ln}\:\left({x}−\mathrm{1}\right)}…
Question Number 72577 by aliesam last updated on 30/Oct/19 $${solve}\:{the}\:{equation} \\ $$$$ \\ $$$${z}\left({x}+{z}\right)\frac{\partial{z}}{\partial{x}}\:−\:{y}\left({y}+{z}\right)\frac{\partial{z}}{\partial{y}}\:=\:\mathrm{0} \\ $$$$ \\ $$$${where}\:{z}=\sqrt{{y}}\:{when}\:{x}=\mathrm{1} \\ $$ Commented by mind is power…
Question Number 6991 by Tawakalitu. last updated on 05/Aug/16 $${Solve}:\:\:\left(\mathrm{1}\:−\:{x}^{\mathrm{2}} \right)\:\frac{{dy}}{{dx}}\:+\:{xy}\:=\:{xp} \\ $$$$ \\ $$$$ \\ $$ Answered by Yozzii last updated on 05/Aug/16 $${x}\neq\mathrm{1}:\:{y}'+\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 6661 by Tawakalitu. last updated on 09/Jul/16 $${Solve}\:{this}\:{equation}\:{by}\:{reducing}\:{it}\:{from}\:{non}\:{homogeneous} \\ $$$${equation}\:{to}\:{homogeneous}\:{equation} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2}\left({x}\:+\:{y}\:−\:\mathrm{1}\right)}{\mathrm{3}{x}\:+\:{y}\:+\:\mathrm{1}} \\ $$$$ \\ $$$${Please}\:{help}\:{with}\:{this}\:{one}\:{too}. \\ $$$${i}\:{was}\:{trying}\:{your}\:{approah}\:{but}\:{not}\:{the}\:{same} \\ $$$$ \\ $$$${thanks}\:{for}\:{your}\:{help}. \\…