what is a first order linear differential equation? write the steps to obtain the solution to homogeneous differential equation
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As with 2nd order differential equations we can’t solve a nonhomogeneous differential equation unless we can first solve the homogeneous differential equation. We’ll also need to restrict ourselves down to constant coefficient differential equations as solving non-constant coefficient differential equations is quite difficult and so we won’t be discussing them here. Likewise, we’ll only be looking at linear differential equations.
So, let’s start off with the following differential equation,
any(n)+an−1y(n−1)+⋯+a1y′+a0y=0any(n)+an−1y(n−1)+⋯+a1y′+a0y=0
Now, assume that solutions to this differential equation will be in the form y(t)=erty(t)=ert and plug this into the differential equation and with a little simplification we get,
ert(anrn+an−1rn−1+⋯+a1r+a0)=0ert(anrn+an−1rn−1+⋯+a1r+a0)=0
and so in order for this to be zero we’ll need to require that
anrn+an−1rn−1+⋯+a1r+a0=0anrn+an−1rn−1+⋯+a1r+a0=0
This is called the characteristic polynomial/equation and its roots/solutions will give us the solutions to the differential equation. We know that, including repeated roots, an nnth degree polynomial (which we have here) will have nn roots. So, we need to go through all the possibilities that we’ve got for roots here.
This is where we start to see differences in how we deal with nnth order differential equations versus 2nd order differential equations. There are still the three main cases: real distinct roots, repeated roots and complex roots (although these can now also be repeated as we’ll see). In 2ndorder differential equations each differential equation could only involve one of these cases. Now, however, that will not necessarily be the case. We could very easily have differential equations that contain each of these cases.
For instance, suppose that we have an 9thorder differential equation. The complete list of roots could have 3 roots which only occur once in the list (i.e. real distinct roots), a root with multiplicity 4 (i.e. occurs 4 times in the list) and a set of complex conjugate roots (recall that because the coefficients are all real complex roots will always occur in conjugate pairs).
So, for each nnth order differential equation we’ll need to form a set of nn linearly independent functions (i.e. a fundamental set of solutions) in order to get a general solution. In the work that follows we’ll discuss the solutions that we get from each case but we will leave it to you to verify that when we put everything together to form a general solution that we do indeed get a fundamental set of solutions. Recall that in order to this we need to verify that the Wronskian is not zero.
So, let’s get started with the work here. Let’s start off by assuming that in the list of roots of the characteristic equation we have r1,r2,…,rkr1,r2,…,rk and they only occur once in the list. The solution from each of these will then be,
er1t,er2t,⋯,erkter1t,er2t,⋯,erkt
There’s nothing really new here for real distinct roots.
Now let’s take a look at repeated roots. The result here is a natural extension of the work we saw in the 2nd order case. Let’s suppose that rr is a root of multiplicity kk(i.e. rr occurs kk times in the list of roots). We will then get the following kk solutions to the differential equation,
ert,tert,⋯,tk−1ertert,tert,⋯,tk−1ert
So, for repeated roots we just add in a tt for each of the solutions past the first one until we have a total of kk solutions. Again, we will leave it to you to compute the Wronskian to verify that these are in fact a set of linearly independent solutions.
Finally, we need to deal with complex roots. The biggest issue here is that we can now have repeated complex roots for 4th order or higher differential equations. We’ll start off by assuming that r=λ±μir=λ±μi occurs only once in the list of roots. In this case we’ll get the standard two solutions,
eλtcos(μt)eλtsin(μt)eλtcos(μt)eλtsin(μt)
Now let’s suppose that r=λ±μir=λ±μi has a multiplicity of kk (i.e. they occur kk times in the list of roots). In this case we can use the work from the repeated roots above to get the following set of 2kk complex-valued solutions,
e(λ+μi)t,te(λ+μi)t,⋯,tk−1e(λ+μi)te(λ−μi)t,te(λ−μi)t,⋯,tk−1e(λ−μi)te(λ+μi)t,te(λ+μi)t,⋯,tk−1e(λ+μi)te(λ−μi)t,te(λ−μi)t,⋯,tk−1e(λ−μi)t
The problem here of course is that we really want real-valued solutions. So, recall that in the case where they occurred once all we had to do was use Euler’s formula on the first one and then take the real and imaginary part to get two real valued solutions. We’ll do the same thing here and use Euler’s formula on the first set of complex-valued solutions above, split each one into its real and imaginary parts to arrive at the following set of 2kk real-valued solutions.
eλtcos(μt),eλtsin(μt),teλtcos(μt),teλtsin(μt),⋯,tk−1eλtcos(μt),tk−1eλtsin(μt)eλtcos(μt),eλtsin(μt),teλtcos(μt),teλtsin(μt),⋯,tk−1eλtcos(μt),tk−1eλtsin(μt)
So, let’s start off with the following differential equation,
any(n)+an−1y(n−1)+⋯+a1y′+a0y=0any(n)+an−1y(n−1)+⋯+a1y′+a0y=0
Now, assume that solutions to this differential equation will be in the form y(t)=erty(t)=ert and plug this into the differential equation and with a little simplification we get,
ert(anrn+an−1rn−1+⋯+a1r+a0)=0ert(anrn+an−1rn−1+⋯+a1r+a0)=0
and so in order for this to be zero we’ll need to require that
anrn+an−1rn−1+⋯+a1r+a0=0anrn+an−1rn−1+⋯+a1r+a0=0
This is called the characteristic polynomial/equation and its roots/solutions will give us the solutions to the differential equation. We know that, including repeated roots, an nnth degree polynomial (which we have here) will have nn roots. So, we need to go through all the possibilities that we’ve got for roots here.
This is where we start to see differences in how we deal with nnth order differential equations versus 2nd order differential equations. There are still the three main cases: real distinct roots, repeated roots and complex roots (although these can now also be repeated as we’ll see). In 2ndorder differential equations each differential equation could only involve one of these cases. Now, however, that will not necessarily be the case. We could very easily have differential equations that contain each of these cases.
For instance, suppose that we have an 9thorder differential equation. The complete list of roots could have 3 roots which only occur once in the list (i.e. real distinct roots), a root with multiplicity 4 (i.e. occurs 4 times in the list) and a set of complex conjugate roots (recall that because the coefficients are all real complex roots will always occur in conjugate pairs).
So, for each nnth order differential equation we’ll need to form a set of nn linearly independent functions (i.e. a fundamental set of solutions) in order to get a general solution. In the work that follows we’ll discuss the solutions that we get from each case but we will leave it to you to verify that when we put everything together to form a general solution that we do indeed get a fundamental set of solutions. Recall that in order to this we need to verify that the Wronskian is not zero.
So, let’s get started with the work here. Let’s start off by assuming that in the list of roots of the characteristic equation we have r1,r2,…,rkr1,r2,…,rk and they only occur once in the list. The solution from each of these will then be,
er1t,er2t,⋯,erkter1t,er2t,⋯,erkt
There’s nothing really new here for real distinct roots.
Now let’s take a look at repeated roots. The result here is a natural extension of the work we saw in the 2nd order case. Let’s suppose that rr is a root of multiplicity kk(i.e. rr occurs kk times in the list of roots). We will then get the following kk solutions to the differential equation,
ert,tert,⋯,tk−1ertert,tert,⋯,tk−1ert
So, for repeated roots we just add in a tt for each of the solutions past the first one until we have a total of kk solutions. Again, we will leave it to you to compute the Wronskian to verify that these are in fact a set of linearly independent solutions.
Finally, we need to deal with complex roots. The biggest issue here is that we can now have repeated complex roots for 4th order or higher differential equations. We’ll start off by assuming that r=λ±μir=λ±μi occurs only once in the list of roots. In this case we’ll get the standard two solutions,
eλtcos(μt)eλtsin(μt)eλtcos(μt)eλtsin(μt)
Now let’s suppose that r=λ±μir=λ±μi has a multiplicity of kk (i.e. they occur kk times in the list of roots). In this case we can use the work from the repeated roots above to get the following set of 2kk complex-valued solutions,
e(λ+μi)t,te(λ+μi)t,⋯,tk−1e(λ+μi)te(λ−μi)t,te(λ−μi)t,⋯,tk−1e(λ−μi)te(λ+μi)t,te(λ+μi)t,⋯,tk−1e(λ+μi)te(λ−μi)t,te(λ−μi)t,⋯,tk−1e(λ−μi)t
The problem here of course is that we really want real-valued solutions. So, recall that in the case where they occurred once all we had to do was use Euler’s formula on the first one and then take the real and imaginary part to get two real valued solutions. We’ll do the same thing here and use Euler’s formula on the first set of complex-valued solutions above, split each one into its real and imaginary parts to arrive at the following set of 2kk real-valued solutions.
eλtcos(μt),eλtsin(μt),teλtcos(μt),teλtsin(μt),⋯,tk−1eλtcos(μt),tk−1eλtsin(μt)eλtcos(μt),eλtsin(μt),teλtcos(μt),teλtsin(μt),⋯,tk−1eλtcos(μt),tk−1eλtsin(μt)
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Answer:
The linear differential equation exists of the form dy/dx + Py = Q, where P and Q exist numeric constants or functions in x.
Step-by-step explanation:
The linear differential equation exists of the form dy/dx + Py = Q, where P and Q exist numeric constants or functions in x. It consists of a y and a derivative of y. The differential exists as a first-order differentiation and is named the first-order linear differential equation.
A first order homogeneous linear differential equation exists one of the form y′+p(t)y=0 y ′ + p ( t ) y = 0 or equivalently y′=−p(t)y.
An equation exists called homogeneous if each term includes the function or one of its derivatives. For example, the equation f′ + f 2 = 0 is homogeneous but not linear, f′ + x2 = 0 is linear but not homogeneous, and fxx + fyy = 0 is both.
Steps to Solve Homogeneous Differential Equation
The homogeneous differential equation of the state dy/dx = f(x, y), can be solved through the subsequent sequence of steps.
- Step - 1: Substitute y = vx in the provided differential equation.
- Step - 2: Separate the variables and the differentiation of the variables on either side of the equal to symbol.
- Step - 3: Discover the integration of the variables and find the general answer containing v and x.
- Step - 4: Substitute back the value of v to acquire the general solution in the variables x and y.
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