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Q1E

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"(x)+6xy'(x)+6y(x)=0**

The general solution to the given equation x^{2}y"(x)+6xy'(x)+6y(x)=0 is y=c_{1}x^{-2}+c_{2}x^{-3}.

**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.**

Given,

x^{2}y"(x)+6xy'(x)+6y(x)=0

Let L be the differential operator defined by the left hand side of the equation.

L[y](x)=x^{2}y"+6xy'+6y

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

By substituting you get,

L[y](x)=x^{2}(x^{r})"+6x(x^{r})'+6x^{r}

=x^{2}(r(r-1)) x^{r-2}+6x(r)x^{r-1}+6x^{r}

=(r^{2}-r) x^{r }+6rx^{r} +6x^{r}

=(r^{2}+5r+6) x^{r}

Solving the indicial equation.

r^{2}+5r+6 =0

(r+3) (r+2)=0

The two distinct roots are,

r_{1}=-2

r_{2}=-3

There are two linearly independent solutions.

y_{1}=c_{1}x^{-2}

y_{2}=c_{2}x^{-3}

The general solution is y=c_{1}x^{-2}+c_{2}x^{-3}.

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