Question Number 109400 by mohammad17 last updated on 23/Aug/20 | ||
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Commented by mohammad17 last updated on 23/Aug/20 | ||
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$${help}\:{me}\:{sir}? \\ $$ | ||
Commented by bobhans last updated on 23/Aug/20 | ||
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$${using}\:{Exact}\:{diff}\:{equation} \\ $$ | ||
Commented by mohammad17 last updated on 23/Aug/20 | ||
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$${sir}\:{the}\:{equation}\:{is}\:{not}\:{exact}\:? \\ $$$${are}\:{you}\:{can}\:{help}\:{me}\:? \\ $$ | ||
Commented by mohammad17 last updated on 23/Aug/20 | ||
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$${no}\:{one}\:{here}\:{can}\:{you}\:{solve}\:{this}\:? \\ $$ | ||
Answered by ajfour last updated on 24/Aug/20 | ||
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$$\left(\mathrm{1}+{x}^{\mathrm{2}} {y}\right){dx}+\left({x}^{\mathrm{2}} +\mathrm{1}\right){ydy}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\frac{\mathrm{1}}{{y}}\left(\frac{\mathrm{1}+{x}^{\mathrm{2}} {y}}{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$$....... \\ $$ | ||