Solve the equations $$\dfrac{dy}{dx}= \dfrac{y-x}{y+x}$$

Asked by Sakshi | 1 year ago |  82

##### Solution :-

From the question it is given that, $$(\dfrac{dy}{dx}) =\dfrac{ (y – x)}{(y + x)}$$

The given differential equation is a homogeneous equation,

Let us assume, y = vx and $$(\dfrac{dy}{dx}) = v + x (\dfrac{dv}{dx})$$

$$v + x (\dfrac{dv}{dx})=\dfrac{(vx – x)}{(vx + x)}$$

Then, $$v + x (\dfrac{dv}{dx})=\dfrac{(v – 1)}{(v + 1)}$$

$$x (\dfrac{dv}{dx}) = \dfrac{(v – 1)}{(v + 1)} – v$$

$$x (\dfrac{dv}{dx}) = \dfrac{(v – 1 – v^2 – v)}{(v + 1)}$$

On dividing we get,

$$x (\dfrac{dv}{dx}) =\dfrac{ – (1 + v^2)}{(v + 1)}$$

Now, taking like variables on same side,

$$\dfrac{(v + 1)}{(v^2 + 1)} = -\dfrac{dx}{x}$$

Integrating on both side we get,

$$\int \dfrac{(v + 1)}{(v^2 + 1)}dv = -\int \dfrac{dx}{x}$$

Then,

$$log \dfrac{(y^2 + x^2)}{x^2}+ 2 tan^{-1} (\dfrac{y}{x}) = 2 log (\dfrac{c}{x})$$

log [x2 + y2] – 2 log x + 2 tan-1 ($$\dfrac{y}{x}$$) = 2 log ($$\dfrac{c}{x}$$)

log (x2 + y2) + 2 tan-1 ($$\dfrac{y}{x}$$) = 2 log c

log (x2 + y2) + 2 tan-1 ($$\dfrac{y}{x}$$) = k

Answered by Aaryan | 1 year ago

### Related Questions

#### Find the particular solution satisfying the given condition (dy/dx) = (x – 1)dy/dx = 2x3y

Find the particular solution satisfying the given condition $$(\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y$$

#### If the interest is compounded continuously at 6% per annum, how much worth Rs. 1000 will be after 10 years?

If the interest is compounded continuously at 6% per annum, how much worth Rs. 1000 will be after 10 years? How long will it take to double Rs. 1000?

#### In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours.

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?