Question Number 96138 by john santu last updated on 30/May/20
$${xy}'\:+\:{y}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} {e}^{{x}} \: \\ $$
Commented by Sourav mridha last updated on 30/May/20
![it′s wrong,it should be xy^′ +y=x^2 e^x . then ans is−− xy=x^2 e^x −2[xe^x −e^x ]+c](Q96172.png)
$$\boldsymbol{{it}}'\mathrm{s}\:\mathrm{wrong},\mathrm{it}\:\boldsymbol{{should}}\:\boldsymbol{{be}} \\ $$$$\boldsymbol{{xy}}^{'} +\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\boldsymbol{{x}}} . \\ $$$$\boldsymbol{{then}}\:\boldsymbol{{ans}}\:\boldsymbol{{is}}−− \\ $$$$\boldsymbol{{xy}}=\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{2}\left[\boldsymbol{{xe}}^{\boldsymbol{{x}}} −\boldsymbol{{e}}^{\boldsymbol{{x}}} \right]+\boldsymbol{{c}} \\ $$