Differential equations: Solution methods for linear-second order ODEs
Solving linear second-order ODEs
Here we list what we have learned about solving a linear second-order differential equation.
Solution Method for a linear second-order ODE
The general solution of the linear second-order differential equation \[\frac{\dd^2 y}{\dd t^2}+p(t)\cdot \frac{\dd y}{\dd t}+q(t)\cdot y=g(t)\]
where #p(t)#, #q(t)#, #g(t)# are continuous functions, can often be found by using the following three steps.
The starting point is the theorem Linear structure of linear ODEs. This reduces the problem of finding the general solution to that of finding two linearly independent solutions #y_1#, #y_2# of the homogeneous equation \( y''+p(x)\cdot y'+q(x)\cdot y=0\) and one (particular) solution #y_{\text{part}}# of the original equation. For then the general solution is \[y=y_{\text{part}}+A\cdot y_1+B\cdot y_2\] where #A# and #B# are integration constants.
- Find a homogeneous solution #y_1#. If the coefficients #p(t)# and #q(t)# are constants, then the method of the theorem General solution of a homogeneous linear second-order ODE with constant coefficients can be applied. Otherwise there is no guarantee, but we can try an appropriate Ansatz.
- Find a second homogeneous solution #y_2#. If the coefficients #p(t)# and #q(t)# are constants, then the method of the theorem General solution of a homogeneous linear second-order ODE with constant coefficients gives a solution #y_2# that is linearly independent of #y_1#. Otherwise, we can use Order reduction of a linear second-order ODE for a given solution in which we may choose the Wronskian for the variation of constants.
- Find a particular solution #y_{\text{part}}#. If two linearly independent homogeneous solutions #y_1# and #y_2# are known, then Variation of constants always gives a solution by reduction to linear first-order equations. But this method is cumbersome. A suitable Ansatz can sometimes give a faster result.
There are more special cases. For example, if the variable #y# itself does not occur, then the ODE can be seen as a linear equation of first order in #y'#.
The solution is determined in four steps.
Step 1: Determine a solution #y_1# distinct from #0# of the associated homogeneous differential equation.
Step 2: Determine a second solution #y_2# of the homogeneous differential equation that is linearly independent of #y_1#.
Step 3: Determine a particular solution #y_{\text{part}} # of the differential equation.
Step 4: Write the general solution #y(t)=A \cdot y_1+B \cdot y_2 +y_{\text{part}} #.
Step 1 The corresponding homogeneous differential equation is
\[ t^2\cdot {\it \frac{\dd^2}{\dd x^2} y}-3\cdot t\cdot {\it \frac{\dd }{\dd x} }+4\cdot y =0 \]
We take as Ansatz:
\[y_1(t)=a\cdot t^3+b\cdot t^2+c\cdot t+d\]
Computation of its derivatives gives
\[y'(t) = 3\cdot a\cdot t^2+2\cdot b\cdot t+c \quad \text { and } \quad y''(t) = 6\cdot a\cdot t+2\cdot b \]
Substituting the Ansatz in the homogeneous differential equations, we find:
\[a\cdot t^3+c\cdot t+4\cdot d\]
Comparing the coefficients on the left and right side of the equation, we find:
\[ a=0 ,\ c=0 ,\ d=0 \] Putting the constant(s) # b # equals to #1#, we find the solution #y_1#: \[y_1(t)=t^2\]
Step 2 Write the corresponding homogeneous differential equation in standard form:
\[\frac{\dd^2 }{\dd^2 t} y-{{3}\over{t}} \cdot \frac{\dd }{\dd t} y+ {{4}\over{t^2}} \cdot y=0\]
According to Second solution of a homogeneous linear second-order ODE, we can find a second solution #y_2(t) = c(t) \cdot y_1(t)#, which is linearly independent of #y_1#, of the form
\[\begin{array}{rcl}
c(t) &=& \displaystyle \int \frac{\e^{ -P(t) }}{y_1(t)^2} \dd t\\
\end{array}\]
where #P(t)# is an antiderivative of #p(t)= -{{3}\over{t}}#. Thus we find #P(t)=-3\cdot \ln \left(t\right)#. Substituting
\[
\e^{-P(t)}=t^3 \qquad \text{ and } \qquad y_1(t)=t^2
\]
in the function rule for #c# gives
\[\begin{array}{rcl}
c(t) &=&\displaystyle \int \frac{t^3}{\left(t^2\right)^2} \dd t \\
&=&\displaystyle \int {{1}\over{t}} \, \dd t \\
&=&\displaystyle \ln \left(t\right) \\
\end{array}\]
Accordingly, \(y_2(t)=y_1(t)\cdot c(t)=t^2\cdot \ln \left(t\right)\).
\[y_{\text{part}} =e\cdot t^3+f\cdot t^2+g\cdot t+h \]
Its first and second derivative are
\[y'(t) = 3\cdot e\cdot t^2+2\cdot f\cdot t+g \quad \text { and } \quad y''(t) = 6\cdot e\cdot t+2\cdot f \]
Substituting these function rules into the differential equation gives
\[\left(e-2\right)\cdot t^3+g\cdot t+4\cdot h=0\]
Equating the coefficients on the left and right side of the equation, we find:
\[ e=2 , g=0 , h=0 \] If we choose \( f \) to be equal to #0#, we find the particular solution: \[y_{\text{part}} (t)=2\cdot t^3\]
Step 4 We conclude that the general solution is \[y(t)=A \cdot t^2 + B \cdot t^2\cdot \ln \left(t\right) + 2\cdot t^3\]
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