# Proof of the Transform of Derivatives

For this post, I wanted to upload an interesting proof of the Laplace transform of Derivatives that is used for transforming a given differential equation in the unknown function $f(t)$ into an algebraic equation in $F(s)$. The Theorem is stated as follows: Suppose that the function $f(t)$ is continuous and piecewise smooth for $t \geq 0$ and is of exponential order as $t \to \infty$, so that there exist nonnegative constants $M$, $c$, and $T$ such that $|f(t)| \leq Me^{ct}$ for $t \geq T$. Then $\mathcal{L}\{f'(t)\}$ exists for $s>c$, and $\mathcal{L}\{f'(t)\}=s\mathcal{L}\{f(t)\}-f(0)=sF(s) - f(0)$ Now for the proof of the Theorem in the general case in which $f'$ is merely piecewise continuous.  To do this it must be proven that the limit $\displaystyle \lim_{b\to\infty} \int_{0}^{b}e^{-st}f'(t) dt$ exists and we also need to find its value. With $b$ fixed, let $t_{1},t_{2},...,t_{k-1}$ be the points interior to the interval $[0,b]$ at which $f'$ is discontinuous. Let $t_{0} = 0$ and $t_{k} = b$. Then the interval can be integrated by parts on each interval $(t_{n-1},t_{n})$ where $f'$ is continuous. This yields $\int_{0}^{b} e^{-st}f'(t)dt = \displaystyle \sum_{n=1}^{k} \int_{t_{n-1}}^{t_{n}}e^{-st}f'(t)dt$ and is equivalent to $\displaystyle \sum^{k}_{n=1}[e^{-st}f(t)]^{t_{n}}_{t_{n-1}} + \displaystyle \sum^{k}_{n=1}s \int^{t_{n}}_{t_{n-1}}e^{-st} f(t) dt$, which I will denote as equation $!$. Now the first summation in equation $!$ is equal to $\displaystyle \sum_{n=1}^{k}[e^{-st}f(t)]^{t_{n}}_{t_{n-1}} = [-f(t_{0})+e^{-st_{1}}]+[-e^{-st_{1}}f(t_{1})+e^{-st_{2}}f(t_{2})]+...+[-e^{st_{k-2}}f(t_{k-2})+e^{-st_{k-1}}f(t_{k-1})] +[-e^{st_{k-1}}f(t_{k-1})+e^{-st_{k}}f(t_{k})]$ and telescopes down to $-f(t_{0}) + e^{-st_{k}}f(t_{k}) = -f(0) +e^{-sb}f(b)$. The second summation adds up to $s$ times the integral from $t_{0} = 0$ to $t_{k} = b$. Therefore equation $!$ reduces to $\int_{0}^{b}e^{-st}f'(t) dt = -f(0) + e^{-sb}f(b) +s\int^{b}_{0}e^{-st}f(t) dt$ From the Theorem above I obtain the expression $|e^{-sb}f(b)| \leq e^{-sb}$ where $Me^{cb}=Me^{-b(s-c)} \Rightarrow 0$ if $s>c$. Therefore, finally taking the limits as $b \Rightarrow \infty$ in the preceding equation, we get the desired result $\mathcal{L}\{f'(t)\} = s \mathcal\{f(t)\}-f(0)$

# Velocity/Acceleration Proof

I want to assume that a body moving with velocity $v$ encounters resistance of the form $\frac{dv}{dt} = -kv^{\frac{3}{2}}$ and show that $v(t) = \frac{4v_{0}}{{(kt\sqrt{v_{0}}+2})^2}$ and that $x(t) = x_{0} + \frac{2}{k}\sqrt{v_{0}}(1-\frac{2}{kt\sqrt{v_{0}}+2})$

Let me first explain what the given equation represents. It states the acceleration of the body is the rate of the velocity per rate of time and that is proportional to $v^{\frac{3}{2}}$. The $-k$ on the right hand side of the differential equation means that there is some constant $k$ that represents a “resisting” force that is opposite to the direction of the velocity. Now to solve the differential equation.

Since the equation is separable it can broken apart into the following equation $v^{\frac{-3}{2}} dv = -k dt$. Since I now have both an integrand and a differential element, I can sum up all of the differential elements by integrating which will look like $\int v^{\frac{-3}{2}} dv = \int -k dt$. After integration is complete, I obtain the equation, $-2v^{\frac{-1}{2}} = -kt - C$. After dividing the equation by $-2$ and cross multiplying, I obtain the formula $v^{\frac{1}{2}} = \frac{2}{kt + C}$. Now, I can square both sides of the equation and solve for $v(t)$. This formula will look like $v(t) = \frac{4}{(kt+C)^2}$.  At time $t=0$, the velocity is equal to the initial velocity, $v_{0}$. When I plug in these conditions to the equation, and take the positive root since velocity is positive, I can solve for the constant $C$, to find that $C = \frac{2}{\sqrt{v_{0}}}$. I can then plug this equation into my derived velocity equation to obtain $v(t) = \frac{4}{(kt+\frac{2}{\sqrt{v_{0}}})^2}$ After some minor algebraic modifications, the equation becomes $v(t) = \frac{4v_{0}}{{(kt\sqrt{v_{0}}+2})^2}$ which is what I was seeking to show.

From this, I can write the final equation above as $\frac{dx}{dt} = \frac{4v_{0}}{(kt\sqrt{v_{0}}+2)^2}$. Like the given differential equation, this equation can be separated and integrated which will look as follows$\frac{1}{4v_{0}} \int 1 dx = \int (kt\sqrt{v_{0}} + 2)^{-2} dt$. Using $u$ substitution, I can let $u = (kt\sqrt{v_{0}}+2$. This then implies that $du = k\sqrt{v_{0}} dt$. After making the substitution for $u$ I must multiply the integral by the constant $\frac{k\sqrt{v_{0}}}{k\sqrt{v_{0}}}$, and bring the constant in the numerator into the integral I can now carry out the $u$ substitution for $du$ and integrate which will give me the formula $\frac{x}{4v_{0}} = \frac{-u^{-1}}{k\sqrt{v_{0}}}+C$. To find the constant $C$ I notice that at time $t = 0$, the position is $x_{0}$. After making this substitution for $t$ and $x$ I find that $C=\frac{x_{0}k+2\sqrt{v_{0}}}{4kv_{0}}$. I can then plug this into the equation that I plugged the initial conditions into, and this will give me $\frac{x}{4v_{0}} = \frac{-({kt\sqrt{v_{0}}+2)}^{-1}}{k\sqrt{v_{0}}} +\frac{x_{0}k+2\sqrt{v_{0}}}{4kv_{0}}$. Finally I can find a common denominator and cancel common terms to obtain the the equation $x(t) = \frac{-4\sqrt{v_{0}}}{k^{2}t\sqrt{v_{0}}+2k} + x_{0} + \frac{2\sqrt{v_{0}}}{k}$. Then I can factor out $\frac{2\sqrt{v_{0}}}{k}$ from two of the terms to obtain the final formula $x_{0} + \frac{2\sqrt{v_{0}}}{k} (1- \frac{2}{kt\sqrt{v_{0}}+2})$, which is what I was asked to show. Now, if i take $\displaystyle \lim_{t\to\infty}$ I obtain the function $x(t) = x_{0} + \frac{2\sqrt{v_{9}}}{k}$ which implies that the object travels only a finite distance before coming to a stop.

# Differential Equations Proof

Show that every solution curve for the equation $y'=x-y$ approaches the line $y=x-1$ as $x\to +\infty$.

Before showing what is asked for above I would like to show the derivation of a formula that will come in handy for solving this problem. The derived formula is used for finding the solution to a linear first-order equation.

I begin by writing down a linear first-order equation in the form $\frac{dy}{dx} + P(x)y=Q(x)$ on an interval on which the coefficient functions$P(x)$ and $Q(x)$ are continuous. I next multiply each side of the equation by the integrating factor $\rho(x)=e^{\int P(x) dx}$. An integrating factor for a differential equation is a function, $\rho(x,y)$ such that the multiplication of each side of the differential equation by $\rho(x,y)$ yields an equation in which each side is recognizable as a derivative. After the multiplication I obtain the equation $e^{\int P(x) dx}\frac{dy}{dx}+P(x)e^{\int P(x) dx}y=Q(x)e^{\int P(x) dx}$. Now, because $D_{x}[\int P(x) dx]=P(x)$, the left-hand side is the derivative product $y(x)e^{\int P(x) dx}$ so the equation becomes $D_{x}[y(x)e^{\int P(x) dx}]=Q(x)e^{\int P(x) dx }$. Integration of both sides of the equation gives me $y(x)e^{\int P(x) dx}=\int(Q(x)e^{\int P(x) dx})dx + C$. Finally, I can solve for $y(x)$ which will give me the general solution for linear first-order equations which will be useful for the initial proof. The general solution is $y(x)=e^{-\int P(x)dx}[\int (Q(x)e^{\int P(x) dx}) dx +C]$.

Now, for the proof I start by finding the integrating factor. After writing the given equation in the form $\frac{dy}{dx} + P(x)y=Q(x)$ I find that $P(x)$ is $1$ which implies that $\rho =e^{\int 1 dx}$ which equals $e^{x}$. Now, I can multiply each side of the equation by $\rho$ which gives me $e^{x}\frac{dy}{dx} +ye^{x}=xe^{x}$. Because of the product rule of derivates this equation is equivalent to $D_{x}[e^{x}y]=xe^{x}$. After integrating each side of the equation, I obtain $e^{x}y = \int xe^{x} dx$. For the right side of the equation, I am going to have to apply the Integration by Parts technique which is of the formula $uv - \int v du$. To use this formula, I must choose the right values for $u$ and $v$. I choose $u =x$, and $v = e^{x}$ which implies that $du = 1 dx$ and $dv = e^{x}$. Now, I can substitute the appropriate variables into the Integration by Parts formula to obtain the equation $e^{x}y = xe^{x} - \int e^{x} dx$. I now have a final integration to do on the right side of the equation and after doing so I obtain $e^{x}y = xe^{x} - e^{x} + C$. I can then solve for $y(x)$ to obtain $y = x - 1 + \frac{C}{e^{x}}$. Finally, I can take $\displaystyle \lim_{x \to + \infty}$, and in this limit the term $\frac{C}{e^{x}}$ goes to $0$ because the exponential gets larger and larger as $x$ gets larger and larger which causes the fraction to essentially approach $0$. This means that $y(x)$ approaches the line $x - 1$ which is what was asked to prove.