The general solution of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the differential equation: $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ Separating the variables: $\frac{d y}{y^2+y+1} = -\frac{d x}{x^2+x+1}

Vedclass · Accepted Answer

$x+y+1=c(1-x-y-2 x y)$

Vedclass · Answer

(C) Given the differential equation: $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ Separating the variables: $\frac{d y}{y^2+y+1} = -\frac{d x}{x^2+x+1}$ Completing the square in the denominators: $\int \frac{d y}{(y+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = -\int \frac{d x}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}$ Using the formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + C$: $\frac{2}{\sqrt{3}} 	an^{-1}(\frac{2y+1}{\sqrt{3}}) = -\frac{2}{\sqrt{3}} 	an^{-1}(\frac{2x+1}{\sqrt{3}}) + C_1$ Dividing by $\frac{2}{\sqrt{3}}$ and rearranging: $	an^{-1}(\frac{2y+1}{\sqrt{3}}) + 	an^{-1}(\frac{2x+1}{\sqrt{3}}) = C_2$ Using the identity $	an^{-1} A + 	an^{-1} B = 	an^{-1}(\frac{A+B}{1-AB})$: $	an^{-1} \left[ \frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - (\frac{2y+1}{\sqrt{3}})(\frac{2x+1}{\sqrt{3}})} ight] = C_2$ Simplifying the expression inside: $\frac{\frac{2(x+y+1)}{\sqrt{3}}}{\frac{3 - (4xy+2x+2y+1)}{3}} = 	an C_2$ $\frac{2(x+y+1)}{\sqrt{3}} \cdot \frac{3}{2(1-x-y-2xy)} = 	an C_2$ $\frac{\sqrt{3}(x+y+1)}{1-x-y-2xy} = 	an C_2$ Thus,$x+y+1 = c(1-x-y-2xy)$,where $c = \frac{	an C_2}{\sqrt{3}}$.

The general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ is

Similar Questions

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x$,when $x = 0, y = 1$ is .....

The equation of the curve passing through $(3, 9)$ which satisfies the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$ is

The general solution of the differential equation $x dy - y dx = 0$ represents

The solution of the differential equation $(1 + x^2)(1 + y)dy + (1 + x)(1 + y^2)dx = 0$ is

The general solution of the differential equation $(x-y^{2}) dx + y(5x+y^{2}) dy = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module