Advertisements
Advertisements
Question
Solve `dy/dx = (x+y+1)/(x+y-1) when x = 2/3 and y = 1/3`
Solution
`dy/dx = (x+y+1)/(x+y-1)` ...(i)
Put x + y = t ...(ii)
∴ y = t - x
Differentiating w.r.t.x, we get
∴ `dy/dx = dt/dx -1` ...(iii)
Substituting (ii) and (iii) in (i), we get
`dt/dx -1 = (t+1)/(t-1)`
∴ `dt/dx =(t+1)/(t-1) + 1 = (t+1+t-1)/(t-1)`
∴ `dt/dx = (2t)/(t-1)`
∴ `((t-1)/t)dt = 2dx`
∴ `(1-1/t)dt = 2dx`
Integrating on both sides, we get
`int (1-1/t) dt = 2 int dx`
∴ t - log | t | = 2x + c
∴ x + y -log |x + y| = 2x + c
x - 2x + y - log (x + y) = c
y - x - log (x + y) = c
y - x - c = log (x + y) ...(i)
Putting `x = 2/ 3 and y = 1/ 3` , we get
∴ from (i)
`1/3 - 2/3 - c = log (2/3 + 1/3)`
`(-1)/3 - c = log` (i)
`(-1)/3 - c = 0`
∴ c = `(-1)/3`
Put c = `1/ 3` in (i)
∴ c = `1/3`
`y - x - (-1/3)` = log (x + y)
`y - x +1/3` = log (x + y)
APPEARS IN
RELATED QUESTIONS
For the differential equation, find the general solution:
`dy/dx = (1 - cos x)/(1+cos x)`
For the differential equation, find the general solution:
`dy/dx + y = 1(y != 1)`
For the differential equation, find the general solution:
y log y dx - x dy = 0
For the differential equation, find the general solution:
`x^5 dy/dx = - y^5`
For the differential equation find a particular solution satisfying the given condition:
`cos (dx/dy) = a(a in R); y = 1` when x = 0
Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e x sin x.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, -3). Find the equation of the curve given that it passes through (-2, 1).
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2 y dy = 0, give that `y = pi/4` when x = 0
Solve
y dx – x dy = −log x dx
Solve
`y log y dy/dx + x – log y = 0`
Find the solution of `"dy"/"dx"` = 2y–x.
Solve the differential equation `"dy"/"dx" + 1` = ex + y.
Solve: (x + y)(dx – dy) = dx + dy. [Hint: Substitute x + y = z after seperating dx and dy]
Find the equation of the curve passing through the (0, –2) given that at any point (x, y) on the curve the product of the slope of its tangent and y-co-ordinate of the point is equal to the x-co-ordinate of the point.
Solve the following differential equation
x2y dx – (x3 + y3)dy = 0
The solution of the differential equation, `(dy)/(dx)` = (x – y)2, when y (1) = 1, is ______.
