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Q5E

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.**

**d ^{2}y/dx^{2}=5/x dy/dx-13/x^{2} y**

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

d^{2}y/dx^{2}=5/x dy/dx-13/x^{2} y

This can be re-written in the required form as

x^{2}y"-5xy'+13y=0

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

L[y](x) = x^{2}y"-5xy'+13y

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} )"-5x(x^{r})'+13(x^{r})

=x^{2} (r(r-1)) x^{r-2} -5x(r) x^{r-1} +13x^{r}

=(r^{2}-r) x^{r} -5rx^{r}+13x^{r}

=(r^{2}-6r+13) x^{r}

Solving the indicial equation,

r^{2}-6r+13=0

r=[2± √(36-4**×**13)]/2

=(6± 4i)/2

=3±2i

There are two complex conjugates,

r_{1}=3+2i and r_{2}=3-2i

Thus there are two linearly independent solutions given by,

x^{3+2i}=x^{3} cos(2 lnx)+ix^{3} sin (2 lnx)

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

y_{1}=x^{3} cos(2 lnx) and y_{2}=x^{3} sin(2 lnx)

Therefore, the general solution for the equation will be,

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

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