Question Number 71570 by TawaTawa last updated on 17/Oct/19 $$\mathrm{Solve}:\:\:\:\:\:\left(\mathrm{2x}\:+\:\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}''\:−\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{x}\right)\mathrm{y}'\:+\:\mathrm{2y}\:\:=\:\:\mathrm{0} \\ $$$$\mathrm{Given}\:\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$ Answered by mind is power last updated on 17/Oct/19…
Question Number 5991 by sanusihammed last updated on 08/Jun/16 $${Solve}\:{the}\:{differential}\:{equation}\: \\ $$$$ \\ $$$${cos}^{\mathrm{2}} \left({x}\right)\:\frac{{dy}}{{dx}}\:+\:{y}\:=\:{tan}\left({x}\right) \\ $$ Answered by prakash jain last updated on 09/Jun/16…
Question Number 71516 by oyemi kemewari last updated on 16/Oct/19 $$\mathrm{5y}''=\left(\mathrm{1}+\mathrm{y}'^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{please}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{differtial}\:\mathrm{equation} \\ $$ Answered by mind is power last updated on 17/Oct/19…
Question Number 137027 by liberty last updated on 29/Mar/21 $$\mathrm{2x}\:\mathrm{dx}\:+\:\mathrm{x}^{−\mathrm{2}} \left(\mathrm{x}\:\mathrm{dy}−\mathrm{y}\:\mathrm{dx}\right)\:=\:\mathrm{0} \\ $$ Answered by bobhans last updated on 29/Mar/21 Terms of Service Privacy Policy…
Question Number 5908 by sanusihammed last updated on 04/Jun/16 $${Solve}\:\: \\ $$$$\frac{{dy}}{{dx}}\:\:−\:{y}\:{tan}\:{x}\:\:=\:{y}^{\mathrm{2}} \:{tan}^{\mathrm{2}} \:{x} \\ $$$$ \\ $$$${please}\:{help}. \\ $$ Commented by 123456 last updated…
Question Number 136815 by rs4089 last updated on 26/Mar/21 Answered by Dwaipayan Shikari last updated on 26/Mar/21 $${y}={e}^{\lambda{x}} \\ $$$${x}^{\mathrm{2}} \lambda^{\mathrm{2}} +{x}\lambda+\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{−{x}\pm\sqrt{{x}^{\mathrm{2}}…
Question Number 5587 by sanusihammed last updated on 21/May/16 $${Please}\:{i}\:{need}\:{your}\:{help}. \\ $$$$ \\ $$$${if}\:\:\:{y}\:\:=\:\:\left[{tanx}\right]^{\left[{tanx}\right]^{\left[{tanx}\right]} } \:\:\:.\:\:\:{find}\:\:{dy}/{dx}\:\:{at}\:\:\Pi/\mathrm{4} \\ $$ Commented by prakash jain last updated on…
Question Number 5208 by sanusihammed last updated on 30/Apr/16 $${If}\:{y}\:=\:{x}!\: \\ $$$$ \\ $$$${Find}\:\frac{{dy}}{{dx}} \\ $$ Commented by FilupSmith last updated on 30/Apr/16 $${x}!=\Gamma\left({x}+\mathrm{1}\right) \\…
Question Number 70720 by oyemi kemewari last updated on 07/Oct/19 $$\mathrm{y}''+\frac{\mathrm{y}'}{\mathrm{x}}+\frac{\mathrm{4y}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{1}+\mathrm{x}^{\mathrm{2}} \\ $$ Commented by Rasheed.Sindhi last updated on 07/Oct/19 $$\mathcal{T}{hank}\:{you}! \\ $$ Commented…
Question Number 70694 by oyemi kemewari last updated on 07/Oct/19 Answered by mind is power last updated on 07/Oct/19 $${let}\:\frac{{y}'}{{x}}={u} \\ $$$$\Rightarrow{y}'={xu}\Rightarrow{y}''={u}+{xu}' \\ $$$${xy}''={xu}+{x}^{\mathrm{2}} {u}'…