Reducing Order of ODE Ly with Ansatz y2: Find u to Solve

samleemc
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Ly ≡ (x +1)^2y′′− 4(x +1)y′+6y =0

given y[1]=(x+1)^2 is a solution, use the ansatz y2(x)= u(x)(x+1)2 to reduce
the order of the differential equation and find a second independent solution y2

how to reduce !? and i can't find u ...can't solve (x+1)^2u''+6u=0

please help!
thx!
 
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Welcome to PF!

Hi samleemc! Welcome to PF! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)

Put y = u(x) (x+1)2 into the original equation …

what do you get? :smile:​
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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