Question

show that the set of functions{x, 1/x} forms a basis of the differential equation x^2y^"+xy^'-y=0 obtain aprticular solutiin when y(1)=1,y(1)=2
​

Answer 1

Answer:

We can let y=xm, so we have: y′(x)=mxm−1,  y′′(x)=m(m−1)xm−2

.

Substituting this back into (1)

, yields:

x2y′′+y=x2(m(m−1)xm−2)+xm=xm(m2−m+1)=0

.

So, we have a characteristic equation:

m2−m+1→m1,2=12±i3–√2

Now, because we are given x>0

(if not, all the x

terms would have absolute values), we can write:

y(x)=y1(x)+y2(x)=c1xm1+c2xm2=c1x12+i3√2+c2x12−i3√2

Using Euler's identity, some algebra, and the fact that x>0

, we can write this as:

y(x)=c1x−−√cos(3–√2lnx)+c2x−−√sin(3–√2lnx

Step-by-step explanation: