Find particular solution of the differential equation:
x^2 dy = 2xy + y^2 dx where y=1 when x=1

Open in App

Solution

Dear Student,

$\frac{dy}{dx} = \frac{2 xy + y^{2}}{x^{2}} . . . . (1) Put y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} Now, (1) becomes, v + x \frac{dv}{dx} = \frac{2 {vx}^{2} + v^{2} x^{2}}{x^{2}} \Rightarrow v + x \frac{dv}{dx} = 2 v + v^{2} \Rightarrow x \frac{dv}{dx} = v + v^{2} \Rightarrow \frac{dv}{v^{2} + v} = \frac{dx}{x} \Rightarrow \int \frac{dv}{v^{2} + v} = \int \frac{dx}{x} \Rightarrow \int \frac{dv}{v^{2} + v + \frac{1}{4} - \frac{1}{4}} = \log |x| + C \Rightarrow \int \frac{dv}{{(v + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} = \log |x| + C \Rightarrow \frac{1}{2 \times 1 / 2} . \log |\frac{v + \frac{1}{2} - \frac{1}{2}}{v + \frac{1}{2} + \frac{1}{2}}| = \log |x| + C \Rightarrow \log |\frac{v}{v + 1}| = \log |x| + C \Rightarrow \log |\frac{y / x}{y / x + 1}| = \log |x| + C \Rightarrow \log |\frac{y}{x + y}| = \log |x| + C . . . . (2) Put x = 1; y = 1, we get \log |\frac{1}{1 + 1}| = \log 1 + C \Rightarrow \log (\frac{1}{2}) = C \Rightarrow C = - \log 2 Now, from (2), we get \log |\frac{y}{x + y}| = \log |x| - \log 2 \log |\frac{y}{x + y}| = \log |\frac{x}{2}| \Rightarrow \frac{y}{x + y} = \frac{x}{2} \Rightarrow 2 y = x^{2} + xy$

Regards

Suggest Corrections

Similar questions

\frac{d x}{d y} = 1 + x^{2} + y^{2} + x^{2} y^{2}

y = 1

x = 0.

x^{2} d y = (2 x y + y^{2}) d x,

y = 1

x = 1

Find the particular solution of the differential equation $x^{2} d y = (2 x y + y^{2}) d x,$ given that y= 1 when x = 1.

Find the particular solution of the differential equation $(1 + x^{2}) \frac{d y}{d x} = (e^{m t a n^{- 1} x} - y),$ given that y =1 when x = 0.

(1 - y^{2}) (1 + log x) d x + 2 x y d y = 0

y = 0

x = 1

Show that the general solution of the differential equation is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter

Join BYJU'S Learning Program