Solve the differential equations condition $$(x + y)^2 \dfrac{dy}{dx} = 1$$

Asked by Sakshi | 1 year ago |  58

##### Solution :-

From the question it is given that, $$(x + y)^2 (\dfrac{dy}{dx}) = 1$$ … [equation (i)]

Let us assume that, x + y = v

Differentiating with respect to x on both side, we get,

$$1 + (\dfrac{dy}{dx}) = \dfrac{dv}{dx}$$

Transposing,

$$\dfrac{dy}{dx} = (\dfrac{dy}{dx}) – 1$$… [equation (ii)]

Substituting equation (ii) in equation (i),

Then, $$v^2 ((\dfrac{dv}{dx}) – 1) = 1$$

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

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

Now, taking like variables on same side,

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

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

Integrating on both side we get,

$$∫\dfrac{v^2 + 1 – 1}{v^2 + 1}dv = ∫dx$$

$$= \int \dfrac{1-1}{v^2+1}=∫dx$$

v – tan-1 v = x + c

x + y – tan-1 (x + y) = x + c

y – tan-1 (x + c) = c

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?