Question about a Pertubation Problem: y''+vy^(1/2)+8=x

Shurukin

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Apr 17, 2018
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I've been asked to solve the differential equation below with perturbation methods

y''+vy^(1/2)+8=x

where 0<v<<1 (ie. what is usually represented by epsilon)

I know to start by supposing that y=a0 + a1v + a2v^2+...
But I'm lost as to what I should do about the free-standing x, keeping in mind that y is a function of x; I also don't know how exactly the y being to the power of a half will change the perturbation series that I should be trying to replace y with. If someone could clarify these two issues, I should hopefully be able to take the rest on myself.
 
Start by replacing \(\displaystyle y^{1/2}\) with its McLaurin series. That will give you integer powers of v.
 
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