Answer to Question #138206 in Differential Equations for Nikhil Singh

Question

Solve
p^2 -2xyp+4y^2=0,where p=dy/dx

Answer 1

Given,

$$p^2 -2xyp+4y^2=0$$

Now, solve the quadratic equation in p, thus we get,

$$p=\frac{2xy\pm \sqrt{4x^2y^2-16y^2}}{2}\\ \implies p=y(x\pm\sqrt{x^2-4})$$

Hence,

$$\frac{dy}{dx}=y(x\pm\sqrt{x^2-4})\\ \implies \frac{dy}{y}=(x\pm\sqrt{x^2-4})dx\\ \implies \int \frac{dy}{y}=\int(x\pm\sqrt{x^2-4})dx\\ \implies \ln y=\int xdx\pm\int \sqrt{x^2-4})dx\\ \implies \ln y=\frac{x^2}{2}\pm \int\sqrt{x^2-4})dx$$

Since,

we know from standard integral that

Thus,

$$\int \sqrt{x^2-4}dx=\dfrac{x\sqrt{x^2-4}}{2}-2\ln\left(\left|\sqrt{x^2-4}+x\right|\right)+C$$

Hence,

$$\ln y=\frac{x^2}{2}\pm \dfrac{x\sqrt{x^2-4}}{2}-2\ln\left(\left|\sqrt{x^2-4}+x\right|\right)+C\\ \implies y=e^{\frac{x^2}{2}\pm \dfrac{x\sqrt{x^2-4}}{2}-2\ln\left(\left|\sqrt{x^2-4}+x\right|\right)+C}\\ \implies y=C_1e^{\frac{x^2}{2}\pm \dfrac{x\sqrt{x^2-4}}{2}-2\ln\left(\left|\sqrt{x^2-4}+x\right|\right)}$$