Write the equation in the form y\' = f(y/x) then use the substitution y = xu to

Question

Write the equation in the form y' = f(y/x) then use the substitution y = xu to find an implicit general solution. Then solve the initial value problem. y' = 3y + 2x/x, y(1) = 2 The resulting differential equation in x and u can be written as xu' = which is separable. Separating variables we arrive at du = dx/x Integrating both sides and simplifying, the solution can be written in the form u + 1 = Cf(x) where C is an arbitrary constant and f(x) =. Transforming back into the variables x ans y and using the initial condition to find C we find the explicit solution of the initial value problem is y =

Explanation / Answer

y' = (3y + 2x)/x , y(1) = 2

substituting y = xu

==> (xu)' = (3xu + 2x)/x

==> (1)u + xu' = x(3u + 2)/x        ; since (uv)' = u'v + uv' ; d/dx xn = nxn-1 , here n = 1

==> u + xu' = 3u + 2

==> xu' = 3u + 2 - u = 2u + 2

==> xu' = 2(u + 1)

using variable seperable method

==> u'/[2(u + 1)] = 1/x               

==> [1/(2(u + 1))] du = (1/x) dx                 ; u' = du/dx

Integrating on both sides

==> [1/(2(u + 1))] du = (1/x) dx

==> (1/2) ln(u + 1) = lnx + c

==> ln(u + 1) = 2lnx + c

==> u + 1 = e2lnx + c

==> u + 1 = e2lnx * ec

==> u + 1 = eln x^2 * C             ; since bln a = ln ab ; and let ec = C

==> u + 1 = C x2                      ; since elna = a

==> F(x) = x2

y = xu ==> u = y/x

==> (y/x) + 1 = C x2

==> (y + x)/x = C x2

==> y + x = C x3

==> y = C x3 - x

y(1) = 2

==> 2 = C (1) - 1

==> 2 = C - 1

==> C = 3

Hence solution is y = 3x3 - x

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