Solutions of differential equations

Differential equations: Introduction to Differential equations

Solutions of differential equations

The exponential function \[y(t)=\e^t\] is equal to its own derivative and so satisfies the differential equation \(y'=y\). In other words, it is a solution of the ODE. The equation \(y'=y\) has other solutions, for example: \[y(t)= 2\e^t,\quad y(t)= -\e^t,\quad y(t)= -\tfrac{1}{3}\e^t \] These solutions are all of the form #y(t)=C\cdot \e^t# for a certain constant \(C\). Every solution is of this form:

Solution of the exponential growth equation

Let #\lambda# be a real number. The general solution of the differential equation \[\frac{\dd y}{\dd t}=\lambda\cdot y\] is \[y(t)=C\cdot \e^{\lambda\cdot t}\] where \(C\) is a constant.

Suppose that #y# is a solution. In order to prove the theorem we show that #z(t)=y(t)\cdot\e^{-t}# is a constant function.
We have #y(t)= z(t)\cdot \e^t# and, thanks to the product rule for differentiation, \[\begin{array}{rcl}y'(t)&=&\frac{\dd}{\dd t}\left(z(t)\cdot \e^t\right)\\ &=& z'(t)\cdot \e^t + z(t)\cdot \frac{\dd}{\dd t}\bigl(\e^t\bigr)\\ &=& z'(t)\cdot \e^t + z(t)\cdot \e^t\end{array}\] The fact that the function \(y\) satisfies the differential equation \(y'=y\) means \[z'(t)\cdot \e^t +z(t)\cdot \e^t=z(t)\cdot \e^t\] This is equivalent to #z'(t)\cdot \e^t=0#, and since #\e^t# is never zero, to #z'(t)=0#. In view of the Uniqueness of the anti-derivative up to a constant, this means that the function \(z(t)\) is constant. Therefore, there is a constant #C# such that #C=y(t)\cdot\e^{-t}#. This means #y(t)=C\cdot\e^{t}#.

The special case where #\lambda=0# is already known: the only functions having derivatives #0#, are the constant functions.

In general, a differential equation can have multiple solutions, but by using constants, often called integration constants, we can sometimes describe the general solution of a differential equation by means of a function rule containing integration constants.

Initial Value Problem

The existence of global solutions shows that an ODE may have a whole host of solutions. To pinpoint a given solution additional conditions are required. When these conditions are all related to the value of the solution #y# or a derivative thereof for a given value of the independent variable (think of the state #y(0)# at time \(0\) or the speed #y'(0)# at time \(0\)), then we speak of an initial value problem.

If there is a unique solution of the ODE satisfying the initial condition (by this we mean the condition(s) of the initial value problem), we call it the specific solution of the initial value problem (or associated with the initial condition).

Consider the initial value problem \[y'=y\text{ with initial condition }\rv{t,y}=\rv{0,1}\]

The initial condition #\rv{t,y}=\rv{0,1}# means that the graph of the solution #y# should go through the point #\rv{0,1}#; in other words, #y(0)=1#.

The general solution of the ODE, as we saw above, is #y(t)=C\cdot\e^t#. This function satisfies the initial condition if #y(0)=1#, so if #C=1#. Therefore, the specific solution of the initial value problem is \[y(t)=\e^t\]

If all the additional requirements are related to the boundary of an interval on which the function is defined (think of #y(2)=7# and #y(9)=13# for the function #y# on the interval #\ivcc{2}{9}#), then we also speak of a boundary value problem. We also talk about the specific solution of the boundary value problem.

Each specific solution needs to be differentiable. As a consequence, it is also continuous.

Solve the following initial value problem in which #y# is a function of #t#:
\[y'=y,\quad y(1)=6\]

\(y=6\cdot \e^{t-1}\)

The general solution of this differential equation of exponential growth equals \[y= C\cdot \e^t\] We use the condition \(y(1)=6\) by substituting #t=1# and #y=6# into it. This gives the equation \(6=C\cdot \e^{1} \), so #C = 6\cdot \e^ {- 1 }#. Substituting this value of #C# into the general solution, we find the special solution of the initial value problem:
\[ y = 6\cdot \e^ {- 1 } \cdot \e ^ t =6\cdot \e^{t-1} \]

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