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Question Number 196458 by peter frank last updated on 25/Aug/23
Answered by Peace last updated on 25/Aug/23
$$\int\left({sin}\left({y}\right)+{ycos}\left({y}\right)\right){dh}=\int{x}\left(\mathrm{2}{ln}\left(\mathrm{x}\right)+\mathrm{1}\right)\mathrm{dx} \\ $$$$\Leftrightarrow+{ysin}\left({y}\right)+{c}={x}^{\mathrm{2}} {ln}\left({x}\right) \\ $$$${x}^{\mathrm{2}} {ln}\left({x}\right)−{ysin}\left({y}\right)+{c}=\mathrm{0} \\ $$$$\Rightarrow−\frac{\pi}{\mathrm{2}}+{c}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {ln}\left({x}\right)−{ysin}\left({y}\right)+\frac{\pi}{\mathrm{2}}={f}\left({x},{y}\right)\:{solution} \\ $$