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Q3E

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"+xy'(x)+17y=0**

The general solution for the given equation is y=c_{1} x cos (4 lnx)+c_{2} x sin(4 lnx)..

**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"+xy'(x)+17y=0

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

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

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^{2}(x^{r})"-x(x^{r})'+17(x^{r})

=x^{2}^{ }(r(r-1)) x^{r-2}-x(r)x^{r-1}+17x^{r}

=(r^{2}-r) x^{r}-rx^{r} +17x^{r}

=(r^{2}-2r+17) x^{r}

Solving the indicial equation

r^{2}-2r+17=0

r=[2± √(4-4**×17**)]/2

=(2± 8i)/2

=1±4i

There are two complex conjugates,

r_{1}=1+4i and r_{2}=1-4i

Thus there are two linearly independent solutions given by

x^{1+4i}=x^{1} cos(4 ln x)+ix^{1} sin(4 ln x)

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

y_{1}=x cos(4 ln x) and y_{2}=x sin (4 ln x)

The general solution for the equation will be,

y=c_{1} x cos (4 lnx)+c_{2} x sin(4 lnx)

Therefore, the general solution for the given equation is y=c_{1} x cos (4 lnx)+c_{2} x sin(4 lnx).

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