Detailed Explanations for Calculus 2 Equations

I am interested in the following differential equation: $y^{″} - q y$ where $q : R_{+} \to R_{+}^{\cdot}$ is a continuous, positive function.

palmantkf4u 2022-04-01

Solving $y^{(2)} - 5 y^{(1)} + 4 y = \frac{1}{e^{x} + 1}$ .

annlanw09y 2022-04-01

Solving $2 x y^{'} (x - y^{2}) + y^{3} = 0$

Alfredo Holmes 2022-04-01

How to you find the general solution of $\frac{d y}{d x} = x {\cos x}^{2}$ ?

avalg10o 2022-04-01

How to solve this nonlinear diff eq of celestial mechanics?
${(\dot{r})}^{2} = \frac{2 μ}{r} + 2 h$
Where mu and h are constants.
I have no idea how to solve it, maybe there is a trick I didn't know.
The only thing that came in mind is to integrate
$\int \frac{d r}{\sqrt{\frac{2 μ}{r} + 2 h}} = \int d t$
but I don't think this is really a solution, since I don't know too how to evaluate the integral in terms of elementary functions.

nastupnat0hh 2022-04-01

How to solve this ordinary differential equation
$t x^{‴} + 3 x^{″} - t x^{'} - x = 0$ ?
For equation $t x^{‴} + 3 x^{″} - t x^{'} - x = 0$ , we know a special solution $x_{1} = \frac{1}{t}$ , how to general solution?
I firstly attempted $d (t x^{″} + 2 x^{'} - t x) = 0$ , then $t x 2 x^{'} - t x = C$ , C is a constant.But in next step , I found that my solution is wrong.Since $x_{1} = \frac{1}{t}$ is a special solution of $t x^{‴} + 3 x^{″} - t x^{'} - x = 0$ , we found $x_{2} = - \frac{1}{t}$ is a solution of equation.Then $x = x_{1} - x_{2} = \frac{2}{t}$ is a solution of $t x^{″} + 2 x^{'} - t x = 0$ . As you can see ,the step is wrong.
Then I attemped other way to solve this equation ,but all failed.Could help me solve this equation?

abitinomaq1 2022-04-01

How to solve
${(y^{'})}^{4} x - 2 y {(y^{'})}^{3} + 12 x^{3} = 0$

loraliyeruxi 2022-04-01

Question on a differential equation
Let $x : R \to R$ a solution of the differenzial equation:
$5 x^{″} (t) + 10 x^{'} (t) + 6 x (t) = 0$
Proof that the function:
$f : ℝ \to ℝ$

London Douglas 2022-04-01

Prove instability using Lyapunov function
$x^{'} = x^{3} + x y$
$y^{'} = - y + y^{2} + x y - x^{3}$

Taylor series of $\frac{x^{2}}{2} - 1 + \cos x$ using power series operations

nastupnat0hh 2022-03-31

I have two problems for which I know the answers (and working) but am still confused about the method used to solve them. The equations are $y y^{(2)} = {(y^{(1)})}^{2} and x^{(2)} + {(x^{(1)})}^{2} = 0$
In my notes it instructs that in the case of the independent variable missing from the equation (x and t, respectively), the substitution to reduce the order is made as follows:
$p = y^{(1)}$
$y^{(2)} = \frac{d p}{d x} = \frac{d p}{d y} \frac{d y}{d x} = p \frac{d p}{d y}$
Then, for the first equation:
$y y^{(2)} = {(y^{(1)})}^{2} \Rightarrow y p \frac{d p}{d y} = p^{2}$
But for the second equation, the substitution is made as:
$x^{(2)} + {(x^{(1)})}^{2} = 0 \Rightarrow \frac{d p}{d t} = - p^{2}$
I've tested this with online calculators and even they compute these two differential equations in the two different ways.
Any explanation of where I'm going wrong would be greatly appreciated.

amonitas3zeb 2022-03-31

I have the differential equation
$d \frac{d}{d x} (p (x) d \frac{d f}{d x}) + \dots = 0$
and I want to perform a generic change of variable from x to $y = y (x)$ .

Oliver Carson 2022-03-31

Specific Solution to Differential Equation
Solve $x {(x - 1)}^{2} g^{″} (x) + (x - 1) g^{'} (x) - x g (x) = \frac{x - 1}{x}, x \geq 1 .$
I already know the general solutions to this as
$g (x) = c_{1} P (x) + c_{2} Q (x) + g_{s} (x) .$
Here, $P (x) = x P_{α}^{2} (2 x - 1), Q (x) = x Q_{α}^{2} (2 x - 1), α = \frac{\sqrt{5} - 1}{2}$
P and Q are the associated Legendre functions and $g_{s} (x)$ is a special solution to this equation. I also know that the special solution can be determined by an integral
$g_{s} (x) = 2 Q (x) \int \frac{P (x)}{x^{3}} d x - 2 P (x) \int \frac{Q (x)}{x^{3}} d x$ .
Can anyone help me find a particular solution to this equation?

Jaylin Clements 2022-03-31

Solving summation problem
$\sum_{i = x}^{2 x} 5 - \sum_{k = 2 x + 1}^{5 x + 2} 5 = - 25$

nomadzkia0re 2022-03-31

Solving separable ODEs
$\frac{d x}{d t} = f (x) g (t)$ with $x (0) = x_{0}$

Octavio Chen 2022-03-31

Solving Shifted Data IVP with Laplace transform
${\begin{cases} y^{″} + y = 2 t \\ y (\frac{π}{4}) = \frac{π}{2} \\ y^{'} (\frac{π}{4}) = 2 - \sqrt{(} 2) \end{cases}$

tibukooinm 2022-03-31

How to solve the ODE
$y' = \frac{x - e^{x}}{x + e^{y}}$

loraliyeruxi 2022-03-31

How to solve $X_{x x} = (1 + δ {(x)}) X$
I am trying to find the vibration modes of a string that has a uniform mass density, plus some point mass somewhere attached to it, modelled by an additional Dirac delta function in the mass density. The wave equation is of the form
$u_{x x} = (1 + δ (x)) u_{t t}$
where u is the deformation, and $(1 + δ (x))$ the mass density. After separation of variables we find
$X_{x x} = (1 + δ (x)) X$
where X is the spatial part of the solution. Is there any analytical solution for X?

amonitas3zeb 2022-03-31

How to solve $K x'' (t) = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] x (t)$
What I'm thinking is to consider the first order version
$x' (t) = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] x (t)$
which I know how to solve, the solution is
$x (t) = c [\begin{matrix} e^{- t} \\ e^{- t} \end{matrix}]$
How do I use this to solve the second-order equation?

ICESSRAIPCELOtblt 2022-03-31

How to solve homogenous differential equation $(x + y)^{2} d x = x y d y$ ?
First i find value of $\frac{d y}{d x} = \frac{{(x + y)}^{2}}{x} y t (i)$
let $y = v x$
differentiating both sides: $\frac{d y}{d x} = v + x d \frac{v}{d x} (ii)$
I tried to solve using equation (i) and (ii) but I am stuck.
${(x + y)}^{2} x = x y d y$

When you are dealing with any Calculus 2 homework, it is vital to have a look at the various questions and answers that will help you see whether you are correct in your approach to finding solutions. Even if you are dealing with analytical aspects of Calculus 2, it will be helpful as you are looking at provided equations and learn how the answers relate to original questions and problems specified.

Do not be afraid to take a look at the basic integration and related application if Calculus 2 does not sound clear or start with the Calculus 1 first.

Calculus 2

Calculus 2 Equations: Expert Solutions and More