Consider the differential equation 2t^2y′′+3ty′−y=0, t>0 . Let y(t) = v(t)y1(t), where v(t) is a function?
I need help solving this differential equation. Let y(t) = v(t)y1(t), where v(t) is a function we wish to solve. Substitute this y into the differential equation, simplify, and rewrite the result as a first-order differential equation. Solve that first-order differential equation
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Let y = v z (where we use z instead of y1 for simplicity). Then
y ' = v ' z + v z' and y '' = v'' z + 2 v' z' + v z'' so, replacing in the original equation we get, after simplification
(1) 2 t^2 z v '' + (2 t^2 z ' + 3t z) v ' + (2 t^2 z'' + 3t z' - z) v = 0
Let us now make z = t^a where a is to be determined. Then we have
z ' = a t^(a-1) and z'' = a(a-1) t^a-2 and therefore
2 t^2 z'' + 3t z' - z = [ 2 a^2 + a - 1 ] t^a so let us choose a such that
2 a^2 + a - 1 = 0 which means
a = ( -1 +/- sqrt (1+8))/4 i.e. a= -1 or a= 1/2
Let us take a = -1, i.e. z=1/t . Then equation (1) above becomes
2t v'' + v' = 0. Put u=v'. Then we have
2t du = -u dt or
2 (du/u) = -(dt/t) which integrates to
u = (dv/dt) = 2k t^(-1/2) and so
v = k t^(1/2) + C and therefore y = v z = k t^(-1/2) + (C/t)
Let us now do the case a = 1/2. Then (1) becomes
t v'' + 2 v' = 0 or t u' = -2u or
du/u = -2 dt/t which integrates to
u = dv/dt = - k t^(-2) and so
v = (k/t) + C and so y = v z = k t^(-1/2) + C t^(1/2)