The general linear first-order differential equation has the following form. \[a_1(t)\cdot y'(t)+a_0(t)\cdot y(t)+b(t)=0\] Because the order of the ODE equals #1#, the function #a_1(t)# cannot be the constant function #0#. Without too great a loss of generality (we will sometimes have to limit ourselves to a smaller domain for the functions in question), we can divide the terms of the equation by #a_1(t)#, so the equation attains the form #y'+p(t)\cdot y = q(t)#, which we will call the standard form of a linear first-order ODE. We show how to find the solution of this ODE.
Suppose that #p(t)# and #q(t)# are continuous functions and that \(P(t)\) is an antiderivative of \(p(t)\). Then the general solution of the ODE
\[y'+p(t)\cdot y = q(t)\] is equal to \[y(t)=\e^{-P(t)}\cdot(F(t)+C)\] where #C# is a constant and #F(t)# is an antiderivative of \(\e^{P(t)}\cdot q(t)\).
In case of the initial value \(y(t_0)=y_0\), the specific solution corresponds to the choice \[C= \e^{P(t_0)}\cdot y_0 - F(t_0)\]
The function #y_{\text{part}} = \e^{-P(t)}\cdot F(t)# is a particular solution and the function #y_{\text{hom}} = C\cdot \e^{-P(t)}# runs through all homogeneous solutions for varying #C#.
To prepare for the proof of the statement we multiply all terms of the ODE by #\e^{P(t)}#, where #P(t)# is an antiderivative of #p(t)#. The idea is that then we can collect the terms #y'# and #p(t)\cdot y# into the single derivative #z'(t)#, where #z(t) =\e^{P( t)}\cdot y#. The factor #\e^{P(t)}# is called an integrating factor.
By use of this integrating factor, we can write the homogeneous solutions as \( y_{\text{part}}(t) = \e^{-P(t)}\cdot C\), where #C# is an integration constant. All that is needed to find the general solution is to find a particular solution #y_{\text{part}}#, for then
\[ y (t ) = y_{\text{part}}+\e^{-P(t)}\cdot C \]
We seek an integrating factor #g(t)#. To this end we multiply all terms of the equation by this factor.
\[\begin{array}{rcl} g(t)\cdot y'+g(t)\cdot p(t)\cdot y &=& g(t)\cdot q(t)\\ &&\phantom{xx}\color{blue}{\text{multiplied by the as yet unknown}}\\ &&\phantom{xx}\color{blue}{\text{integrating factor }g(t)}\\ g(t)\cdot y'+g'(t) \cdot y &=& g(t)\cdot q(t)\quad \text{ and } \quad g'(t)=g(t)\cdot p(t)\\ &&\phantom{xx}\color{blue}{g(t)\text{ can be found since } g'= g\cdot p(t)\text{ is separable}}\\ \frac{\dd}{\dd t}\left(g(t)\cdot y\right)&=& g(t)\cdot q(t)\quad\text{ and }\quad g(t)=\e^{P(t)}\\ &&\phantom{xx}\color{blue}{\text{product rule for differentiation }}\\ &&\phantom{xx}\color{blue}{g(t)=\e^{P(t)} \text{ is a solution of }g'= g\cdot p(t)} \\\e^{P(t)}\cdot y &=&\displaystyle\int\e^{P(t)}\cdot q(t)\,\dd t\\ &&\phantom{xx}\color{blue}{\text{antiderivative calculated and }g(t)=\e^{P(t)}\text{ substituted}}\\ y &=&\displaystyle\e^{-P(t)} \cdot \int\e^{P(t)}\cdot q(t)\,\dd t\\ &&\phantom{xx}\color{blue}{\text{divided by }\e^{P(t)}}\\ y &=&\e^{-P(t)} \cdot \left(F(t)+C\right)\\ &&\phantom{xx}\color{blue}{\text{antiderivative }F(t)\text{ of }\e^{P(t)}\cdot q(t)\text{ used}} \end{array}\]
The function #g(t)=\e^{P(t)}# thus appears to be a good integration factor.
The theorem suggests a general method for finding the solution: first find an antiderivative #P(t)# of #p(t)# and then an antiderivative #F(t)# of \(\e^{P(t)}\cdot q(t)\). By use of these functions, the solution can be written as #y(t)=\e^{-P(t)}\cdot(F(t)+C)#, where #C# is an integration constant.
Often there are other methods, such as separation of variables if #q(t)# is the product of a function of #y# and a function of #t#, or finding a particular solution and the solving the homogeneous equation.
Solve the following initial value problem. \[\frac{\dd y}{\dd t}-3y=5\e^{4 t},\qquad y(0)=6\]
#y(t)=## 5 \euler^{4 t }+\euler^{3 t }#
Multiplying both sides of the ODE by the integrating factor \(\e^{-3 t}\) yields \[\e^{-3 t}\, y'-3 y\cdot\e^{-3 t}=5\e^{4 t}\cdot\e^{-3 t}\] We apply the product rule for differential forms to the left-hand side and rewrite it as a single derivative. \[\frac{\dd\left(y\cdot\e^{-3 t}\right)}{\dd t}=5\e^t\]Integrating the left and right side gives \[y\cdot\e^{-3 t}=5\e^t+C\] where #C# is the integration constant. It follows that the general solution is \[y=5\e^{4 t}+C\cdot\e^{3 t}\] Substitution of the initial value \(y(0)=6\) leads to the equation \(5+C=6\), whose solution is \(C=1\). Thus, the solution of the initial value problem is \[y(t)=5\e^{4 t}+\e^{3 t}\]
Using the
Chain rule for differentiation, we check that #y(t)=5\e^{4t}+\e^{3 t}# indeed satisfies the differential equation: \[\frac{\dd}{\dd t}y(t)
=\frac{\dd}{\dd t}\left(5\e^{4t}+\e^{3 t}\right)
=5\e^{4t}\cdot 4+\e^{3 t}\cdot 3
=3 \cdot\left(5\e^{4t}+\e^{3 t}\right)+5\cdot\e^{4 t}
=3 y+5\cdot\e^{4 t}
\]
It is also easy to verify that #y(0)=6#:
\[y(0)=5\e^{4\cdot 0}+\e^{3 \cdot 0}=5+1 = 6\]
In the figure below, the slope field of the ODE is drawn. The initial value can be found as the red point #\rv{0,6}# at the left. The integral curve from this point to the point #\rv{1,5\cdot \euler^4+\euler^3}#, the red point at the right, is drawn blue.
Solving linear first-order ODEs
