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Q7E

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 ^{3}y"'+4x^{2}y"+10xy'-10y=0**

The general solution for the given equation is y=c_{1}x+c_{2}x^{-1} cos(3 ln x)+c_{3} x^{-1} sin(3 ln x).

**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^{3}y"'+4x^{2}y"+10xy'-10y=0

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

L[y](x)=x^{3}y"'+4x^{2}y"+10xy'-10y

And let's set

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

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

L[y](x)=x^{3}(x^{r})"'+4x^{2}(x^{r})"+10x(x^{r})'-10(x^{r})

=x^{3} (r(r-1)(r-2)) x^{r-3} +4x^{2} (r(r-1)) x^{r-2} +10x(r) x^{r-1}-10x^{r}

=(r^{3}-3r^{2}+2r) x^{r}+4 (r^{2}-r) x^{r} +10r x^{r}-10x^{r}

=(r^{3}+r^{2}+8r-10) x^{r}

Solving the indicial equation,

r^{3}+r^{2}+8r-10=0

By substitution, we can identify that r=1 is a solution of the equation,

(r-1) (r^{2}+2r+10)=(r-1) [r- (-2± √(4-4** ×10**)]/2)]

=(r-1) [r- (-2±6i)/2]

=(r-1) [r-(-1±3i)]

There is one real root and two complex conjugates,

r_{1}=1

r_{2}= -1+3i

r_{3}= -1-3i

For complex roots there are two linearly independent solutions given by,

x^{1+3i }= x^{-1} cos (3 ln x)+ix^{-1} sin(3 ln x)

You can write three linearly independent real-valued solutions as,

y_{1}=x,

y_{2}=x^{-1} cos(3 ln x)

y_{3}=x^{-1} sin(3 ln x)

Therefore, the general solution for the equation will be,

y=c_{1}x+c_{2}x^{-1} cos(3 ln x)+c_{3} x^{-1} sin(3 ln x)

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