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Q4E

Expert-verifiedFound in: Page 450

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In Problems 1-10, use a substitution y=x ^{r} to find the general solution to the given equation for x>0. **

**x ^{2}y"+2xy'-3y=0**

The general solution for the given equation is y=c_{1}x^{-1/2+√13/2} +c_{2}x^{-1/2-√13/2} .

**In mathematics, a Cauchy problem is one in which the solution to a partial differential equation must satisfy specific constraints specified on a hypersurface in the domain. **

**An initial value problem or a boundary value problem is both examples of Cauchy problems. **

**The equation will be in the form of ax ^{2}y"+bxy'+cy=0.**

The given equation is,

x^{2}y"+2xy'-3y=0

Let L be the differential operator defined by the left-hand side of equation, that is

L[y](x)=x^{2}y"+2xy'-3y

Let's see,

w(r,x)=x^{r}

Substituting the w(r,x) in place of y(x), you get

L[y](x)=x^{2}(x^{r} )"+2x(x^{r})'-3(x^{r})

=x^{2} (r(r-1)) x^{r-2}+2x (r) x^{r-1}-3 (x^{r})

=(r^{2}-r) x^{r}+2rx^{r}-3x^{r}

=(r^{2}+r-3) x^{r}

Solving the indicial equation

r^{2}+r-3=0

r= [-1±√(1+12)]/2

r= -1/2±√13/2

There are two distinct roots,

r_{1}= -1/2+√13/2 and r_{2}= -1/2-√13/2

Thus there are two linearly independent solutions given by,

y_{1}=c_{1}x^{-1/2+√13/2} and y_{2}=c_{2}x^{-1/2-√13/2}

Hence, the general solution for the given equation will be,

y=c_{1}x^{-1/2+√13/2} +c_{2}x^{-1/2-√13/2}

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