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Example of an ODE with initial conditions
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I was given the following example as an introduction towards ODEs. But I have an issue with it:
Let $$-u'' + pu' + qu = 0, p,q in mathbbR \
u(t_0) = u_0 \
u'(t_0) = v_0, \ u_0,v_0in mathbbR
$$
If we say, that
$$u = e^lambda(t-t_0) u_0
$$
We fulfill the first initial condition. Then we can write the equation as
$$(-lambda^2 +plambda + q )u = 0$$
If we solve $-lambda^2 +plambda + q = 0$ for $lambda$, we might have either a single or a two-fold zero. Which is way too restricted to let us fulfill the second initial conditon $u'(t_0)=v_0$, because
$$u'(t) = lambda e^lambda (t-t_0)u_0$$ and so $$u'(t_0) = lambda u_0 stackrel!= v_0 Rightarrow lambda = v_0 / u_0$$
But we are already forced to choose $lambda$ from the at most two solutions to the 2nd degree polynomial.
differential-equations examples-counterexamples
asked 1 hour ago
dba
1819
add a comment |Â
up vote
1
down vote
favorite
I was given the following example as an introduction towards ODEs. But I have an issue with it:
Let $$-u'' + pu' + qu = 0, p,q in mathbbR \
u(t_0) = u_0 \
u'(t_0) = v_0, \ u_0,v_0in mathbbR
$$
If we say, that
$$u = e^lambda(t-t_0) u_0
$$
We fulfill the first initial condition. Then we can write the equation as
$$(-lambda^2 +plambda + q )u = 0$$
If we solve $-lambda^2 +plambda + q = 0$ for $lambda$, we might have either a single or a two-fold zero. Which is way too restricted to let us fulfill the second initial conditon $u'(t_0)=v_0$, because
$$u'(t) = lambda e^lambda (t-t_0)u_0$$ and so $$u'(t_0) = lambda u_0 stackrel!= v_0 Rightarrow lambda = v_0 / u_0$$
But we are already forced to choose $lambda$ from the at most two solutions to the 2nd degree polynomial.
differential-equations examples-counterexamples
asked 1 hour ago
dba
1819
2
Usually, you would solve this differential equation by solving for the two values of $lambda$ and then calling them $lambda_1$ and $lambda_2$. Then, you would say $y=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$ and solve for $c_1$, $c_2$ by setting $y(t_0)=u_0$ and $y'(t_0)=v_0$. I have no idea what your textbook here is doing by saying $lambda=fracv_0u_0$.
– Noble Mushtak
57 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was given the following example as an introduction towards ODEs. But I have an issue with it:
Let $$-u'' + pu' + qu = 0, p,q in mathbbR \
u(t_0) = u_0 \
u'(t_0) = v_0, \ u_0,v_0in mathbbR
$$
If we say, that
$$u = e^lambda(t-t_0) u_0
$$
We fulfill the first initial condition. Then we can write the equation as
$$(-lambda^2 +plambda + q )u = 0$$
If we solve $-lambda^2 +plambda + q = 0$ for $lambda$, we might have either a single or a two-fold zero. Which is way too restricted to let us fulfill the second initial conditon $u'(t_0)=v_0$, because
$$u'(t) = lambda e^lambda (t-t_0)u_0$$ and so $$u'(t_0) = lambda u_0 stackrel!= v_0 Rightarrow lambda = v_0 / u_0$$
But we are already forced to choose $lambda$ from the at most two solutions to the 2nd degree polynomial.
differential-equations examples-counterexamples
asked 1 hour ago
dba
1819
I was given the following example as an introduction towards ODEs. But I have an issue with it:
Let $$-u'' + pu' + qu = 0, p,q in mathbbR \
u(t_0) = u_0 \
u'(t_0) = v_0, \ u_0,v_0in mathbbR
$$
If we say, that
$$u = e^lambda(t-t_0) u_0
$$
We fulfill the first initial condition. Then we can write the equation as
$$(-lambda^2 +plambda + q )u = 0$$
If we solve $-lambda^2 +plambda + q = 0$ for $lambda$, we might have either a single or a two-fold zero. Which is way too restricted to let us fulfill the second initial conditon $u'(t_0)=v_0$, because
$$u'(t) = lambda e^lambda (t-t_0)u_0$$ and so $$u'(t_0) = lambda u_0 stackrel!= v_0 Rightarrow lambda = v_0 / u_0$$
But we are already forced to choose $lambda$ from the at most two solutions to the 2nd degree polynomial.
differential-equations examples-counterexamples
differential-equations examples-counterexamples
asked 1 hour ago
dba
1819
asked 1 hour ago
dba
1819
asked 1 hour ago
dba
1819
asked 1 hour ago
dba
1819
asked 1 hour ago
dba
1819
1819
2
Usually, you would solve this differential equation by solving for the two values of $lambda$ and then calling them $lambda_1$ and $lambda_2$. Then, you would say $y=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$ and solve for $c_1$, $c_2$ by setting $y(t_0)=u_0$ and $y'(t_0)=v_0$. I have no idea what your textbook here is doing by saying $lambda=fracv_0u_0$.
– Noble Mushtak
57 mins ago
add a comment |Â
2
Usually, you would solve this differential equation by solving for the two values of $lambda$ and then calling them $lambda_1$ and $lambda_2$. Then, you would say $y=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$ and solve for $c_1$, $c_2$ by setting $y(t_0)=u_0$ and $y'(t_0)=v_0$. I have no idea what your textbook here is doing by saying $lambda=fracv_0u_0$.
– Noble Mushtak
57 mins ago
2
2
Usually, you would solve this differential equation by solving for the two values of $lambda$ and then calling them $lambda_1$ and $lambda_2$. Then, you would say $y=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$ and solve for $c_1$, $c_2$ by setting $y(t_0)=u_0$ and $y'(t_0)=v_0$. I have no idea what your textbook here is doing by saying $lambda=fracv_0u_0$.
– Noble Mushtak
57 mins ago
Usually, you would solve this differential equation by solving for the two values of $lambda$ and then calling them $lambda_1$ and $lambda_2$. Then, you would say $y=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$ and solve for $c_1$, $c_2$ by setting $y(t_0)=u_0$ and $y'(t_0)=v_0$. I have no idea what your textbook here is doing by saying $lambda=fracv_0u_0$.
– Noble Mushtak
57 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
First of all you find the general solution to your differential equation.
You solve the characteristic polynomial and find your eignvalues and form the general solution which is a linear combination, either $$y= c_1 e^lambda _1 t + c_2 e^lambda _2 t$$ in case of distinct eigenvalues or
$$y= c_1 e^lambda t + c_2 te^lambda t$$ in case of an eigenvalue with multiplicity two.
Then you find the parameters $c_1$ and $c_2$ from your initial conditions.
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
add a comment |Â
up vote
2
down vote
Now, I am not sure what your textbook is doing, but I am going to show you how I would solve this equation using the methods I have learned. First, as you said, we have:
$$(-lambda^2+plambda+q)u=0$$
Thus, we need to divide by $u$ and find the solutions to $lambda$. There are then three possibilities, assuming $p$ and $q$ are real numbers.
- There are two, distinct real solutions.
- There is only one solution with multiplicity of two.
- There are two, distinct complex solutions.
In order to make this simpler, I am going to focus on Possibility 1 and I will refer to the two possible values of $lambda$ as $lambda_1$ and $lambda_2$. Now, we have two possible solutions:
$$u=e^lambda_1(t-t_0), u=e^lambda_2(t-t_0)$$
Now, you will learn the reasons for why this is later, but these two solutions are called a "fundamental set of solutions" because they generate all the possible solutions to this ODE. Thus, we write the general solution as a linear combination of these two solutions, like so:
$$u=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$$
Here, $c_1$ and $c_2$ are constants we have to solve for using the initial conditions. As you can see, the problem gives you two initial conditions and there are two constants to solve for, so it works out. First, let's get the equation from $u(t_0)=u_0$:
$$u(t_0)=c_1e^lambda_1(t_0-t_0)+c_2e^lambda_2(t_0-t_0)=c_1+c_2=u_0$$
Now, to get the equation from $u'(t_0)=v_0$, we first have to figure out what $u'$ is. Using chain rule, it is easy to take the derivative of $u$:
$$u'=lambda_1c_1e^lambda_1(t-t_0)+lambda_2c_2e^lambda_2(t-t_0)$$
Thus, from $u'(t_0)=v_0$, we find that:
$$u'(t_0)=lambda_1c_1e^lambda_1(t_0-t_0)+lambda_2c_2e^lambda_2(t_0-t_0)=lambda_1c_1+lambda_2c_2=v_0$$
This, combined with the $c_1+c_2=u_0$ equation, allows us to solve for $c_1$ and $c_2$ since we have a system of two linear equations with two unknowns. Using Cramer's Rule, we get:
$$c_1=frac-u_0lambda_2+v_0lambda_1-lambda_2, c_2=fracu_0lambda_1-v_0lambda_1-lambda_2$$
Thus, our final solution to this ODE with initial conditions is:
$$u=frac-u_0lambda_2+v_0lambda_1-lambda_2e^lambda_1(t-t_0)+fracu_0lambda_1-v_0lambda_1-lambda_2e^lambda_2(t-t_0)$$
Now, this is a pretty long solution and it only covers the case where $lambda_1, lambda_2$ are distinct, real numbers. However, that is the beautiful yet messy world of differential equations for you. I hope this example helps you understand initial condition problems more!
answered 43 mins ago
Noble Mushtak
12.5k1632
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First of all you find the general solution to your differential equation.
You solve the characteristic polynomial and find your eignvalues and form the general solution which is a linear combination, either $$y= c_1 e^lambda _1 t + c_2 e^lambda _2 t$$ in case of distinct eigenvalues or
$$y= c_1 e^lambda t + c_2 te^lambda t$$ in case of an eigenvalue with multiplicity two.
Then you find the parameters $c_1$ and $c_2$ from your initial conditions.
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
add a comment |Â
up vote
3
down vote
accepted
First of all you find the general solution to your differential equation.
You solve the characteristic polynomial and find your eignvalues and form the general solution which is a linear combination, either $$y= c_1 e^lambda _1 t + c_2 e^lambda _2 t$$ in case of distinct eigenvalues or
$$y= c_1 e^lambda t + c_2 te^lambda t$$ in case of an eigenvalue with multiplicity two.
Then you find the parameters $c_1$ and $c_2$ from your initial conditions.
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First of all you find the general solution to your differential equation.
You solve the characteristic polynomial and find your eignvalues and form the general solution which is a linear combination, either $$y= c_1 e^lambda _1 t + c_2 e^lambda _2 t$$ in case of distinct eigenvalues or
$$y= c_1 e^lambda t + c_2 te^lambda t$$ in case of an eigenvalue with multiplicity two.
Then you find the parameters $c_1$ and $c_2$ from your initial conditions.
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
First of all you find the general solution to your differential equation.
You solve the characteristic polynomial and find your eignvalues and form the general solution which is a linear combination, either $$y= c_1 e^lambda _1 t + c_2 e^lambda _2 t$$ in case of distinct eigenvalues or
$$y= c_1 e^lambda t + c_2 te^lambda t$$ in case of an eigenvalue with multiplicity two.
Then you find the parameters $c_1$ and $c_2$ from your initial conditions.
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
answered 49 mins ago
Mohammad Riazi-Kermani
35.5k41855
35.5k41855
add a comment |Â
add a comment |Â
up vote
2
down vote
Now, I am not sure what your textbook is doing, but I am going to show you how I would solve this equation using the methods I have learned. First, as you said, we have:
$$(-lambda^2+plambda+q)u=0$$
Thus, we need to divide by $u$ and find the solutions to $lambda$. There are then three possibilities, assuming $p$ and $q$ are real numbers.
- There are two, distinct real solutions.
- There is only one solution with multiplicity of two.
- There are two, distinct complex solutions.
In order to make this simpler, I am going to focus on Possibility 1 and I will refer to the two possible values of $lambda$ as $lambda_1$ and $lambda_2$. Now, we have two possible solutions:
$$u=e^lambda_1(t-t_0), u=e^lambda_2(t-t_0)$$
Now, you will learn the reasons for why this is later, but these two solutions are called a "fundamental set of solutions" because they generate all the possible solutions to this ODE. Thus, we write the general solution as a linear combination of these two solutions, like so:
$$u=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$$
Here, $c_1$ and $c_2$ are constants we have to solve for using the initial conditions. As you can see, the problem gives you two initial conditions and there are two constants to solve for, so it works out. First, let's get the equation from $u(t_0)=u_0$:
$$u(t_0)=c_1e^lambda_1(t_0-t_0)+c_2e^lambda_2(t_0-t_0)=c_1+c_2=u_0$$
Now, to get the equation from $u'(t_0)=v_0$, we first have to figure out what $u'$ is. Using chain rule, it is easy to take the derivative of $u$:
$$u'=lambda_1c_1e^lambda_1(t-t_0)+lambda_2c_2e^lambda_2(t-t_0)$$
Thus, from $u'(t_0)=v_0$, we find that:
$$u'(t_0)=lambda_1c_1e^lambda_1(t_0-t_0)+lambda_2c_2e^lambda_2(t_0-t_0)=lambda_1c_1+lambda_2c_2=v_0$$
This, combined with the $c_1+c_2=u_0$ equation, allows us to solve for $c_1$ and $c_2$ since we have a system of two linear equations with two unknowns. Using Cramer's Rule, we get:
$$c_1=frac-u_0lambda_2+v_0lambda_1-lambda_2, c_2=fracu_0lambda_1-v_0lambda_1-lambda_2$$
Thus, our final solution to this ODE with initial conditions is:
$$u=frac-u_0lambda_2+v_0lambda_1-lambda_2e^lambda_1(t-t_0)+fracu_0lambda_1-v_0lambda_1-lambda_2e^lambda_2(t-t_0)$$
Now, this is a pretty long solution and it only covers the case where $lambda_1, lambda_2$ are distinct, real numbers. However, that is the beautiful yet messy world of differential equations for you. I hope this example helps you understand initial condition problems more!
answered 43 mins ago
Noble Mushtak
12.5k1632
add a comment |Â
up vote
2
down vote
Now, I am not sure what your textbook is doing, but I am going to show you how I would solve this equation using the methods I have learned. First, as you said, we have:
$$(-lambda^2+plambda+q)u=0$$
Thus, we need to divide by $u$ and find the solutions to $lambda$. There are then three possibilities, assuming $p$ and $q$ are real numbers.
- There are two, distinct real solutions.
- There is only one solution with multiplicity of two.
- There are two, distinct complex solutions.
In order to make this simpler, I am going to focus on Possibility 1 and I will refer to the two possible values of $lambda$ as $lambda_1$ and $lambda_2$. Now, we have two possible solutions:
$$u=e^lambda_1(t-t_0), u=e^lambda_2(t-t_0)$$
Now, you will learn the reasons for why this is later, but these two solutions are called a "fundamental set of solutions" because they generate all the possible solutions to this ODE. Thus, we write the general solution as a linear combination of these two solutions, like so:
$$u=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$$
Here, $c_1$ and $c_2$ are constants we have to solve for using the initial conditions. As you can see, the problem gives you two initial conditions and there are two constants to solve for, so it works out. First, let's get the equation from $u(t_0)=u_0$:
$$u(t_0)=c_1e^lambda_1(t_0-t_0)+c_2e^lambda_2(t_0-t_0)=c_1+c_2=u_0$$
Now, to get the equation from $u'(t_0)=v_0$, we first have to figure out what $u'$ is. Using chain rule, it is easy to take the derivative of $u$:
$$u'=lambda_1c_1e^lambda_1(t-t_0)+lambda_2c_2e^lambda_2(t-t_0)$$
Thus, from $u'(t_0)=v_0$, we find that:
$$u'(t_0)=lambda_1c_1e^lambda_1(t_0-t_0)+lambda_2c_2e^lambda_2(t_0-t_0)=lambda_1c_1+lambda_2c_2=v_0$$
This, combined with the $c_1+c_2=u_0$ equation, allows us to solve for $c_1$ and $c_2$ since we have a system of two linear equations with two unknowns. Using Cramer's Rule, we get:
$$c_1=frac-u_0lambda_2+v_0lambda_1-lambda_2, c_2=fracu_0lambda_1-v_0lambda_1-lambda_2$$
Thus, our final solution to this ODE with initial conditions is:
$$u=frac-u_0lambda_2+v_0lambda_1-lambda_2e^lambda_1(t-t_0)+fracu_0lambda_1-v_0lambda_1-lambda_2e^lambda_2(t-t_0)$$
Now, this is a pretty long solution and it only covers the case where $lambda_1, lambda_2$ are distinct, real numbers. However, that is the beautiful yet messy world of differential equations for you. I hope this example helps you understand initial condition problems more!
answered 43 mins ago
Noble Mushtak
12.5k1632
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Now, I am not sure what your textbook is doing, but I am going to show you how I would solve this equation using the methods I have learned. First, as you said, we have:
$$(-lambda^2+plambda+q)u=0$$
Thus, we need to divide by $u$ and find the solutions to $lambda$. There are then three possibilities, assuming $p$ and $q$ are real numbers.
- There are two, distinct real solutions.
- There is only one solution with multiplicity of two.
- There are two, distinct complex solutions.
In order to make this simpler, I am going to focus on Possibility 1 and I will refer to the two possible values of $lambda$ as $lambda_1$ and $lambda_2$. Now, we have two possible solutions:
$$u=e^lambda_1(t-t_0), u=e^lambda_2(t-t_0)$$
Now, you will learn the reasons for why this is later, but these two solutions are called a "fundamental set of solutions" because they generate all the possible solutions to this ODE. Thus, we write the general solution as a linear combination of these two solutions, like so:
$$u=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$$
Here, $c_1$ and $c_2$ are constants we have to solve for using the initial conditions. As you can see, the problem gives you two initial conditions and there are two constants to solve for, so it works out. First, let's get the equation from $u(t_0)=u_0$:
$$u(t_0)=c_1e^lambda_1(t_0-t_0)+c_2e^lambda_2(t_0-t_0)=c_1+c_2=u_0$$
Now, to get the equation from $u'(t_0)=v_0$, we first have to figure out what $u'$ is. Using chain rule, it is easy to take the derivative of $u$:
$$u'=lambda_1c_1e^lambda_1(t-t_0)+lambda_2c_2e^lambda_2(t-t_0)$$
Thus, from $u'(t_0)=v_0$, we find that:
$$u'(t_0)=lambda_1c_1e^lambda_1(t_0-t_0)+lambda_2c_2e^lambda_2(t_0-t_0)=lambda_1c_1+lambda_2c_2=v_0$$
This, combined with the $c_1+c_2=u_0$ equation, allows us to solve for $c_1$ and $c_2$ since we have a system of two linear equations with two unknowns. Using Cramer's Rule, we get:
$$c_1=frac-u_0lambda_2+v_0lambda_1-lambda_2, c_2=fracu_0lambda_1-v_0lambda_1-lambda_2$$
Thus, our final solution to this ODE with initial conditions is:
$$u=frac-u_0lambda_2+v_0lambda_1-lambda_2e^lambda_1(t-t_0)+fracu_0lambda_1-v_0lambda_1-lambda_2e^lambda_2(t-t_0)$$
Now, this is a pretty long solution and it only covers the case where $lambda_1, lambda_2$ are distinct, real numbers. However, that is the beautiful yet messy world of differential equations for you. I hope this example helps you understand initial condition problems more!
answered 43 mins ago
Noble Mushtak
12.5k1632
Now, I am not sure what your textbook is doing, but I am going to show you how I would solve this equation using the methods I have learned. First, as you said, we have:
$$(-lambda^2+plambda+q)u=0$$
Thus, we need to divide by $u$ and find the solutions to $lambda$. There are then three possibilities, assuming $p$ and $q$ are real numbers.
- There are two, distinct real solutions.
- There is only one solution with multiplicity of two.
- There are two, distinct complex solutions.
In order to make this simpler, I am going to focus on Possibility 1 and I will refer to the two possible values of $lambda$ as $lambda_1$ and $lambda_2$. Now, we have two possible solutions:
$$u=e^lambda_1(t-t_0), u=e^lambda_2(t-t_0)$$
Now, you will learn the reasons for why this is later, but these two solutions are called a "fundamental set of solutions" because they generate all the possible solutions to this ODE. Thus, we write the general solution as a linear combination of these two solutions, like so:
$$u=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$$
Here, $c_1$ and $c_2$ are constants we have to solve for using the initial conditions. As you can see, the problem gives you two initial conditions and there are two constants to solve for, so it works out. First, let's get the equation from $u(t_0)=u_0$:
$$u(t_0)=c_1e^lambda_1(t_0-t_0)+c_2e^lambda_2(t_0-t_0)=c_1+c_2=u_0$$
Now, to get the equation from $u'(t_0)=v_0$, we first have to figure out what $u'$ is. Using chain rule, it is easy to take the derivative of $u$:
$$u'=lambda_1c_1e^lambda_1(t-t_0)+lambda_2c_2e^lambda_2(t-t_0)$$
Thus, from $u'(t_0)=v_0$, we find that:
$$u'(t_0)=lambda_1c_1e^lambda_1(t_0-t_0)+lambda_2c_2e^lambda_2(t_0-t_0)=lambda_1c_1+lambda_2c_2=v_0$$
This, combined with the $c_1+c_2=u_0$ equation, allows us to solve for $c_1$ and $c_2$ since we have a system of two linear equations with two unknowns. Using Cramer's Rule, we get:
$$c_1=frac-u_0lambda_2+v_0lambda_1-lambda_2, c_2=fracu_0lambda_1-v_0lambda_1-lambda_2$$
Thus, our final solution to this ODE with initial conditions is:
$$u=frac-u_0lambda_2+v_0lambda_1-lambda_2e^lambda_1(t-t_0)+fracu_0lambda_1-v_0lambda_1-lambda_2e^lambda_2(t-t_0)$$
Now, this is a pretty long solution and it only covers the case where $lambda_1, lambda_2$ are distinct, real numbers. However, that is the beautiful yet messy world of differential equations for you. I hope this example helps you understand initial condition problems more!
answered 43 mins ago
Noble Mushtak
12.5k1632
answered 43 mins ago
Noble Mushtak
12.5k1632
answered 43 mins ago
Noble Mushtak
12.5k1632
answered 43 mins ago
Noble Mushtak
12.5k1632
12.5k1632
add a comment |Â
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Usually, you would solve this differential equation by solving for the two values of $lambda$ and then calling them $lambda_1$ and $lambda_2$. Then, you would say $y=c_1e^lambda_1(t-t_0)+c_2e^lambda_2(t-t_0)$ and solve for $c_1$, $c_2$ by setting $y(t_0)=u_0$ and $y'(t_0)=v_0$. I have no idea what your textbook here is doing by saying $lambda=fracv_0u_0$.
– Noble Mushtak
57 mins ago