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(1-x)y"-50y'+3y=cosx

(1-x)y"-50y'+3y=cosx

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Answer #1

(1-x)y"-50y'+3y=cos x

considering the homogeneous equation:

This is of the form,

y''+P(x)y'+Q(x)y=0

the normal form is: u''+f(x)u=0

let y(x)=u(x)v(x)

y(x)=u(x)v(x)

so the normal form of the equation is

+f(x)=0

so ,

+=0

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y'-y=cosx
y'-y=cosx
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