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Question-171442




Question Number 171442 by mnjuly1970 last updated on 15/Jun/22
Commented by infinityaction last updated on 15/Jun/22
    e^x (1+x)dx + (ye^y −xe^x )dy = 0      let  t = xe^x       e^x (1+x)dx = dt       dt + (ye^y  − t)dy = 0        (dt/dy) +ye^y  −t = 0       (dt/dy) − t = −ye^y         (dt/dy) + Pt = Q       P  = −1 , Q = −ye^y        I.F  = e^(∫−1dy)   ⇒ I.F = e^(−y)       solution        t×I.F  = ∫(I.F×Q)dy +C       te^(−y)  = ∫(−ye^y ×e^(−y) )dy +C      te^(−y)  + (y^2 /2)  = C       ∴ t =xe^x       xe^((x−y))  + (y^2 /2) = C
$$\:\:\:\:{e}^{{x}} \left(\mathrm{1}+{x}\right){dx}\:+\:\left({ye}^{{y}} −{xe}^{{x}} \right){dy}\:=\:\mathrm{0} \\ $$$$\:\:\:\:{let}\:\:{t}\:=\:{xe}^{{x}} \\ $$$$\:\:\:\:{e}^{{x}} \left(\mathrm{1}+{x}\right){dx}\:=\:{dt} \\ $$$$\:\:\:\:\:{dt}\:+\:\left({ye}^{{y}} \:−\:{t}\right){dy}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\frac{{dt}}{{dy}}\:+{ye}^{{y}} \:−{t}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\frac{{dt}}{{dy}}\:−\:{t}\:=\:−{ye}^{{y}} \\ $$$$\:\:\:\:\:\:\frac{{dt}}{{dy}}\:+\:{Pt}\:=\:{Q} \\ $$$$\:\:\:\:\:{P}\:\:=\:−\mathrm{1}\:,\:{Q}\:=\:−{ye}^{{y}} \\ $$$$\:\:\:\:\:{I}.{F}\:\:=\:{e}^{\int−\mathrm{1}{dy}} \:\:\Rightarrow\:{I}.{F}\:=\:{e}^{−{y}} \\ $$$$\:\:\:\:{solution} \\ $$$$\:\:\:\:\:\:{t}×{I}.{F}\:\:=\:\int\left({I}.{F}×{Q}\right){dy}\:+{C} \\ $$$$\:\:\:\:\:{te}^{−{y}} \:=\:\int\left(−{ye}^{{y}} ×{e}^{−{y}} \right){dy}\:+{C} \\ $$$$\:\:\:\:{te}^{−{y}} \:+\:\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:{C} \\ $$$$\:\:\:\:\:\therefore\:{t}\:={xe}^{{x}} \\ $$$$\:\:\:\:{xe}^{\left({x}−{y}\right)} \:+\:\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\:=\:{C} \\ $$

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