Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 184085 by Shrinava last updated on 02/Jan/23

(y^2 + xy^2 )y^′ + x^2 − yx^2 = 0

$$\left(\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}^{\mathrm{2}} \right)\mathrm{y}^{'} \:+\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{yx}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$

Answered by mr W last updated on 02/Jan/23

((y^2 dy)/(y−1))=((x^2 dx)/(x+1)) ∫((y^2 dy)/(y−1))=∫((x^2 dx)/(x+1)) ∫(((y^2 −1+1)dy)/(y−1))=∫(((x^2 −1+1)dx)/(x+1)) ∫(y+1+(1/(y−1)))dy=∫(x−1+(1/(x+1)))dx ⇒(y^2 /2)+y+ln (y−1)=(x^2 /2)−x+ln (x+1)+C

$$\frac{{y}^{\mathrm{2}} {dy}}{{y}−\mathrm{1}}=\frac{{x}^{\mathrm{2}} {dx}}{{x}+\mathrm{1}} \\ $$$$\int\frac{{y}^{\mathrm{2}} {dy}}{{y}−\mathrm{1}}=\int\frac{{x}^{\mathrm{2}} {dx}}{{x}+\mathrm{1}} \\ $$$$\int\frac{\left({y}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right){dy}}{{y}−\mathrm{1}}=\int\frac{\left({x}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}\right){dx}}{{x}+\mathrm{1}} \\ $$$$\int\left({y}+\mathrm{1}+\frac{\mathrm{1}}{{y}−\mathrm{1}}\right){dy}=\int\left({x}−\mathrm{1}+\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$\Rightarrow\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{y}+\mathrm{ln}\:\left({y}−\mathrm{1}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}+\mathrm{ln}\:\left({x}+\mathrm{1}\right)+{C} \\ $$

Commented by Shrinava last updated on 02/Jan/23

thank you so much professor, cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{professor},\:\mathrm{cool} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com