Solve the initial value problem 2xy dydx z2 3y2SolutionGiv

Solve the initial value problem 2xy dy/dx = (z^2 + 3y^2).

Solution

Given that dy/dx = (x²+3y²)/2xy

   dy/dx = (1/2) [(x/y) + 3(y/x)] ----- Equation (1)

Let y= vx ----Equation (2)

Differentiate on both the sides with respect to x

We know that d/dx (uv) = u (dv/dx) + v (du/dx)

Then, dy/dx = v d/dx (x) + x (dv/dx)  

  dy/dx = v.1 + x (dv/dx)

   dy/dx = v + x (dv/dx)

  From the Equation (1), dy/dx = (1/2) [(x/y) + 3(y/x)]

Therefore, (1/2) [(x/y) + 3(y/x)] = v + x (dv/dx)

From Equation (2) , y=vx

(1/2) [(x/vx) + 3(vx/x)] = v + x (dv/dx)

(1/2) [(1/v) + (3v)] = v + x (dv/dx)

(1/2) [ [(1/v) + (3v) - v ] = x (dv/dx)

(1/2) [(1/v) + (2v)] = x (dv/dx)

On seperating the variabl dv/ es, we wiil get

(1/2) [dx/x] = dv/ [(1/v) + (2v)]

(1/2) [dx/x] = [ v / (1+2v²)] dv

[dx/x] = [ 2v / (1+2v²)] dv

[dx/x] = (1/2) [ 4v / (1+2v²)] dv

Intedgate on both sides

[dx/x] = (1/2) [ 4v / (1+2v²)] dv

We know that Integral of dx/x = In x   and f\'(x)/f(x) dx = In[ f(x)]

Hence,

In x = (1/2) In (1+2v²) + c

From the Equation (2), v = y/x

Therefore,

In x = (1/2) In [1+2(y/x)²] + In c

Inx = (1/2) In [ (x²+ 2y²) /x²] + In c

2 Inx = In [ (x²+ 2y²) /x²] + In c

Inx² = In [ (x²+ 2y²) /x²] + In c

x² = [ (x²+ 2y²) /x²] .c

(x²+ 2y²) c = x⁴

Hence,

the solution is (x²+ 2y²) c = x⁴