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Q15E

Expert-verifiedFound in: Page 453

Book edition
9th

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

Pages
616 pages

ISBN
9780321977069

**In Problems 15-17, solve the given initial value problem t ^{2}x"-12x=0. x(1)=3 and x'(1)=5.**

The solution of the given initial value problem is x=2t^{4}+t^{-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 an example of the Cauchy problem. **

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

The given equation is,

t^{2}x"-12x=0

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

L [x] (t) = t^{2}x"-12x

And let.

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

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

L [w] (t)=t^{2} (t^{r})"-12(t^{r})

=t^{2 }(r (r-1)) t^{r-2}-12(t^{r})

= (r^{2}-r)t^{r}- 12t^{r}

=(r^{2}-r-12)t^{r}

Solving the indicial equation,

r^{2}-r-12=0

(r-4)(r+3)=0

r=4, -3

There are two roots of the above indicial equation.

r_{1}=4 and r_{2}=-3

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

x_{1}=t^{4 }and x_{2}=t^{-3}

The general solution for the equation will be,

x=c_{1}t^{4}+c_{2}t^{-3}

For the given initial conditions.

x(1) = 3 and x'(1)=5

x(t)= c_{1}t^{4}+c_{2}t^{-3}

x(1) = c_{1}+c_{2}

Here,

c_{1}+c_{2} =3

Now,

x'(t)= 4c_{1}t^{3}-4c_{2}t^{-4}

x'(1)= 4c_{1}-3c_{2}

=5

Solving the two simultaneous equations, you get the values of two constants c_{1} and c_{2} as,

c_{1}=2 and c_{2}=1

Thus, the solution of the given initial value problem is,

x=2t^{4}+t^{-3}

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