### 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, \(y_2(t)=y_1(t)\cdot c(t)=t^2\cdot \ln \left(t\right)\).

**Step 3**For finding a particular solution we decide to use an Ansatz. We choose

\[y_{\text{part}} =e\cdot t^3+f\cdot t^2+g\cdot t+h \]

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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