Question

Solve via Laplace transforms y''+9y=1/sqrt(t) if y(0)=0 and y'(0)=1. The convolution theorem might be useful

Answer 1

Okay, let's give it a shot. :-)
$$y''+9y=\frac1{\sqrt{t}}\text{ given }y(0)=0,y'(0)=1$$First, let's begin with taking the Laplace transform \(\mathfrak{L}\{\cdot\}\) of both sides of our equation, where \(\mathfrak{L}\{f(t)\}=\int\limits_0^\infty f(t)e^{-st}\,\mathrm{d}t\).
$$\mathfrak{L}\{y''+9y\}=\mathfrak{L}\left\{\frac1{\sqrt{t}}\right\}$$Since the Laplace transform is linear, we may reduce our left-hand side to \(\mathfrak{L}\{y''\}+9\mathfrak{L}\{y\}\). Let \(Y(s)=\mathfrak{L}\{y(t)\}\) and we can reduce our first term via integration by parts:$$\begin{align*}\int y''(t)e^{-st}\,\mathrm{d}t&=y'(t)e^{-st}+s\int y'(t)e^{-st}\,\mathrm{d}t\\&=y'(t)e^{-st}+s\left(y(t)e^{-st}+s\int y(t)e^{-st}\,\mathrm{d}t\right)\\&=y'(t)e^{-st}+sy(t)e^{-st}+s^2\int y(t)e^{-st}\,\mathrm{d}t\\\mathfrak{L}\{y''\}&=\int_0^\infty y''(t)e^{-st}\,\mathrm{d}t\\&=\lim_{c\to\infty}\int_0^cy''(t)e^{-st}\,\mathrm{d}t\\&=\lim_{c\to\infty}\left(y'(c)e^{-sc}+sy(c)e^{-ct}-y'(0)-sy(0)+s^2\int_0^cy(t)e^{-st}\right)\\&=s^2\int_0^\infty y(t)e^{-st}\,\mathrm{d}t-sy(0)-y'(0)\\&=s^2\mathfrak{L}\{y(t)\}-sy(0)-y'(0)\\&=s^2Y(s)\end{align*}$$

Answer 2

On to our right-hand side, then; first, let's rewrite \(1/\sqrt{t}=t^{-\frac12}\). While I could bother deriving the following transform using integration by parts as well, I'll leave that as an exercise to you.$$\mathfrak{L}\left\{t^{-\frac12}\right\}=\frac{\Gamma(\frac12)}{s^\frac12}=\frac{\sqrt{\pi}}{\sqrt{s}}$$Okay, so our equation is then:$$\begin{align*}s^2Y(s)+9Y(s)=\frac{\sqrt{\pi}}{\sqrt{s}}\\Y(s)(s^2+9)=\frac{\sqrt{\pi}}{\sqrt{s}}\\Y(s)=\sqrt{\pi}\left(\frac1{\sqrt{s}}\cdot\frac1{s^2+9}\right)\end{align*}$$Now we can take the inverse Laplace transform of both sides to yield:$$y(t)=\sqrt{\pi}\mathfrak{L}^{-1}\left\{\frac1{\sqrt{s}}\cdot\frac1{s^2+9}\right\}$$... now here is where the convolution theorem comes into play. Recognize that we're taking the inverse Laplace transform of a product, as the convolution theorem states \(\mathfrak{L}^{-1}\{F(s)\cdot G(s)\}=\mathfrak{L}^{-1}\{F(s)\}*\mathfrak{L}^{-1}\{G(s)\}=f(t)*g(t)\), where \(f(t)*g(t)=\int_{-\infty}^\infty f(t)g(t-\tau)\,\mathrm{d}\tau\), the convolution of \(f(t),g(t)\).

Answer 3

$$y(t)=\left(\mathfrak{L}^{-1}\left\{\frac{\sqrt{\pi}}{s^\frac12}\right\}*\mathfrak{L}^{-1}\left\{\frac1{s^2+9}\right\}\right)$$Let's first tackle our second inverse transform. Recognize $$\frac1{s^2+9}=\frac13\cdot\frac3{s^2+3^2}$$ and thus our inverse is just $$\mathfrak{L}^{-1}\left\{\frac13\cdot\frac3{s^2+3^2}\right\}=\frac13\sin3t$$For our first, we get back what we began with, \(\mathfrak{L}^{-1}\left\{\frac{\sqrt{\pi}}{\sqrt{s}}\right\}=\frac1{\sqrt{t}}\). Thus our solution \(y(t)\) is then just: $$\begin{align*}y(t)&=\frac13\left(\frac1{\sqrt{t}}*\sin3t\right)\\&=\frac13\int_{-\infty}^\infty\frac{\sin3(t-\tau)}{\sqrt{\tau}}\,\mathrm{d}\tau\end{align*}$$

Answer 4

Let's direct our attention to the numerator, \(\sin3(t-\tau)\). We may reduce it as follows:$$\sin3(t-\tau)=\sin(3t-3\tau)=\sin3t\cos3\tau-\sin3\tau\cos3t$$. I think you can take it from here. You should split the integral into two and pull out the \(\sin3t\) and \(\cos3t\) factors as constants. What you're left with you can use a substitution on to yield Fresnel integrals.
http://en.wikipedia.org/wiki/Fresnel_integral

Answer 5

Shouldn't L{y''} = \[s ^{2} Y(s) -1\]?

Answer 6

Since y'(0)=1

Answer 7

Nevermind I think I figured it out. thank you

Answer 8

Long answers like that deserve 5 medals. Nice work.