Question

решить дифференциальное уравнение!

Answer 1

$y'\sqrt{1+y^2}=\dfrac{x^2}{y}\\\\\\\dfrac{dy}{dx}=\dfrac{x^2}{y\sqrt{1+y^2}}\ \ ,\ \ \ \displaystyle \int y\sqrt{1+y^2}\, dy=\int x^2\, dx\ \ ,\\\\\\\star \ \int y\sqrt{1+y^2}\, dy=\Big[\ y=tgt\ ,\ dy=\dfrac{dt}{cos^2t}\ ,\ t=arctgy\ \Big]=\\\\\\=\int tgt\cdot \sqrt{1+tg^2t}\cdot \frac{dt}{cos^2t}=\int tgt\cdot \frac{1}{cost}\cdot \frac{dt}{cos^2t}=\int \frac{sint}{cos^4t}\, dt=\\\\\\=\int \frac{-d\, (cost)}{cos^4t}=\frac{(cost)^{-3}}{-3}+C_1=-\frac{1}{3\, cos^3t}+C_1=-\frac{1}{3\, cos^3(arctgy)}+C_1=$

$=-\dfrac{1}{3\, \Big(\dfrac{1}{\sqrt{1+y^2}}\Big)^3}+C_1=-\dfrac{\sqrt{(1+y^2)^3}}{3}+C_1\ \ \star \\\\\\\\-\dfrac{\sqrt{(1+y^2)^3}}{3}=\dfrac{x^3}{3}+\dfrac{C}{3}\ \ ,\ \ \ \sqrt{(1+y^2)^3}=-x^3-C\ \ ,\ \ 1+y^2=\sqrt[3]{(-x^3-C)^2}\ ,\\\\\\\boxed{\ y^2=1+\sqrt[3]{(x^2+C)^2}\ }$