How do you solve $xy\prime +2y=4{x}^{2}$ given y(1)=0?

Kwenze0l
2022-09-28
Answered

How do you solve $xy\prime +2y=4{x}^{2}$ given y(1)=0?

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Corbin Hanson

Answered 2022-09-29
Author has **10** answers

We have:

$xy\prime +2y=4{x}^{2}$

Which we can write as:

$y\prime +\frac{2}{x}y=4x......\left[1\right]$

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

Then the integrating factor is given by;

$IF={e}^{\int P\left(x\right)dx}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\mathrm{exp}(\int \frac{2}{x}dx)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\mathrm{exp}\left(2\mathrm{ln}x\right)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={e}^{{\mathrm{ln}x}^{2}}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={x}^{2}$

And if we multiply the DE [1] by this Integrating Factor, IF, we will have a perfect product differential;

$y\prime +\frac{2}{x}y=4x$

$\therefore {x}^{2}y\prime +2xy=4{x}^{3}$

$\therefore \frac{d}{dx}\left({x}^{2}y\right)=4{x}^{3}$

Which we can directly integrate to get:

$\int \frac{d}{dx}\left({x}^{2}y\right)dx=\int 4{x}^{3}dx$

$\therefore {x}^{2}y={x}^{4}+C$

Applying the initial condition, y(1)=0, we get:

$1\cdot 0=1+C\Rightarrow C=-1$

Thus the solution is:

${x}^{2}y={x}^{4}-1$

$\therefore y={x}^{2}-\frac{1}{{x}^{2}}$

$xy\prime +2y=4{x}^{2}$

Which we can write as:

$y\prime +\frac{2}{x}y=4x......\left[1\right]$

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

Then the integrating factor is given by;

$IF={e}^{\int P\left(x\right)dx}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\mathrm{exp}(\int \frac{2}{x}dx)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\mathrm{exp}\left(2\mathrm{ln}x\right)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={e}^{{\mathrm{ln}x}^{2}}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={x}^{2}$

And if we multiply the DE [1] by this Integrating Factor, IF, we will have a perfect product differential;

$y\prime +\frac{2}{x}y=4x$

$\therefore {x}^{2}y\prime +2xy=4{x}^{3}$

$\therefore \frac{d}{dx}\left({x}^{2}y\right)=4{x}^{3}$

Which we can directly integrate to get:

$\int \frac{d}{dx}\left({x}^{2}y\right)dx=\int 4{x}^{3}dx$

$\therefore {x}^{2}y={x}^{4}+C$

Applying the initial condition, y(1)=0, we get:

$1\cdot 0=1+C\Rightarrow C=-1$

Thus the solution is:

${x}^{2}y={x}^{4}-1$

$\therefore y={x}^{2}-\frac{1}{{x}^{2}}$

asked 2022-06-15

For a second order differential equation (many physical systems) in one variable, I know "procedures" to compute the energy. Given

${q}^{\u2033}(t)=f(q(t),{q}^{\prime}(t)),\text{}\text{}q(0)={q}_{0},\text{}\text{}{q}^{\prime}(0)={v}_{0},$

if we're lucky we can read off the related Lagrangian L, introduce $p=\frac{\mathrm{\partial}L}{\mathrm{\partial}q}$, do a Legendre transform and we got the Hamiltonian function H(q,p) for which $\frac{\text{d}}{\text{d}t}H(q(t),p(t))=0$ for solutions of the differential equation.

We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=\u27e8q(t),p(t)\u27e9$ in phase space given by a solution with initial conditions X(0), the function H(q,p) always takes the same values. It defines surfaces in phase space indexed by initial conditions X(0).

I wonder how to view this for a priori first order systems

${q}^{\prime}(t)=f(q(t)),\text{}\text{}q(0)={q}_{0},$

where I think this must be even simpler. E.g. I thing some functional

$q(t)\mapsto q(t)-\text{an integral over}\text{}f(q(t))\text{}\text{something},$

should exists which will be constant for solutions of the equation, i.e. only depend on ${q}_{0}$. For each f, the functional dependence of this "energy" on "${q}_{0}$" will be different.

However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?

If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\dot{x}\propto x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.

${q}^{\u2033}(t)=f(q(t),{q}^{\prime}(t)),\text{}\text{}q(0)={q}_{0},\text{}\text{}{q}^{\prime}(0)={v}_{0},$

if we're lucky we can read off the related Lagrangian L, introduce $p=\frac{\mathrm{\partial}L}{\mathrm{\partial}q}$, do a Legendre transform and we got the Hamiltonian function H(q,p) for which $\frac{\text{d}}{\text{d}t}H(q(t),p(t))=0$ for solutions of the differential equation.

We can be more exact and give all the conditions for Noethers theorem to hold and the result is that along the flow $X(t)=\u27e8q(t),p(t)\u27e9$ in phase space given by a solution with initial conditions X(0), the function H(q,p) always takes the same values. It defines surfaces in phase space indexed by initial conditions X(0).

I wonder how to view this for a priori first order systems

${q}^{\prime}(t)=f(q(t)),\text{}\text{}q(0)={q}_{0},$

where I think this must be even simpler. E.g. I thing some functional

$q(t)\mapsto q(t)-\text{an integral over}\text{}f(q(t))\text{}\text{something},$

should exists which will be constant for solutions of the equation, i.e. only depend on ${q}_{0}$. For each f, the functional dependence of this "energy" on "${q}_{0}$" will be different.

However, I can't seem to find a general relation. What's the theory behind this, is there an energy'ish time integral of motion? What is the functional dependence on the intial condition, for a suitable constant of motion for for first order systems. And what would be the interpretation, given that we speak of a situation with only one initial condition?

If it is that case that the system is too restricted so that there is no meaningful geometrical interpretation, then let's think about a system of first order differential equations. This is like the one we generated from phase space, except that it doesn't really come from a second order situation and so the intial conditions aren't really e.g. intial position and velocity. I'm pretty sure there are situation where one considers such a directly generated flow (the equation might be more complicated than exponential flow $\dot{x}\propto x$), but I don't recall any talk about the time-constant of motion in these systems, or how to interpret it.

asked 2022-09-23

Solve the following first-order linear differential equations? $y\prime -4{x}^{3}y=8{x}^{3}$

asked 2022-07-13

I have difficulties to solve these two differential equations:

1) ${y}^{\prime}(x)=\frac{x-y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

but I don't know if it's correct. If it's right, how do I find the constant $c$? Because WolframAlpha says that the solution is $y(x)=\sqrt{2}\sqrt{{x}^{2}+1}-x$.

2) ${y}^{\prime}(x)=\frac{2y(x)-x}{2x-y(x)}$. I'm arrived to prove that $\frac{z-1}{(z+1{)}^{3}}={e}^{2c}{x}^{2}$ but I don't know if it's correct. If it's right, how do I explain $z$ to substitute it in $y=xz$? Then, how do I find the constant $c$ ?

1) ${y}^{\prime}(x)=\frac{x-y(x)}{x+y(x)}$ with the initial condition $y(1)=1$ .I'm arrived to prove that

$y=x(\sqrt{2-{e}^{-2(\mathrm{ln}x+c)}}-1)$

but I don't know if it's correct. If it's right, how do I find the constant $c$? Because WolframAlpha says that the solution is $y(x)=\sqrt{2}\sqrt{{x}^{2}+1}-x$.

2) ${y}^{\prime}(x)=\frac{2y(x)-x}{2x-y(x)}$. I'm arrived to prove that $\frac{z-1}{(z+1{)}^{3}}={e}^{2c}{x}^{2}$ but I don't know if it's correct. If it's right, how do I explain $z$ to substitute it in $y=xz$? Then, how do I find the constant $c$ ?

asked 2022-06-23

I am studying numeric solutions of differential equations, and part of my reading is found in Simmonds' book, Differential Equations with Applications and Historical Notes. Although the chapter on numerical methods is written by John S. Robertson.

In the treatment of Euler's method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?

For some context, the differential equations under consideration are those of the form y'=f(x,y), defined on some interval [a,b], with some initial value $y(a)=\alpha $

These are then transformed into the integral equation

$y({x}_{1})-y({x}_{0})={\int}_{{x}_{0}}^{{x}_{1}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\text{.}$

Then, using Taylor's theorem and substituting back into that formula, the error is found to be

$\frac{{h}^{2}}{2}{y}^{\u2033}(\xi )\text{,}$

which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y''(x) is bounded on the entire interval, and so, by extension, is ${y}^{\u2033}(\xi )$. I don't understand why this is necessarily true.

In the treatment of Euler's method, he states that the second derivative of the solution y is bounded by some constant M. What am I missing that makes it necessary that the solution of a first order differential equation even be twice differentiable?

For some context, the differential equations under consideration are those of the form y'=f(x,y), defined on some interval [a,b], with some initial value $y(a)=\alpha $

These are then transformed into the integral equation

$y({x}_{1})-y({x}_{0})={\int}_{{x}_{0}}^{{x}_{1}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\text{.}$

Then, using Taylor's theorem and substituting back into that formula, the error is found to be

$\frac{{h}^{2}}{2}{y}^{\u2033}(\xi )\text{,}$

which I understand to be true. For the remainder of the method, this quantity is neglected. But in the next section, it simply states that the quantity y''(x) is bounded on the entire interval, and so, by extension, is ${y}^{\u2033}(\xi )$. I don't understand why this is necessarily true.

asked 2022-06-01

Is a first order differential equation categorized by $f({y}^{\prime},y,x)=0$ or ${y}^{\prime}=f(y,x)$?

In the case of the second, why $\mathrm{sin}{y}^{\prime}+3y+x+5=0$ isnt a first order differential equation?

In the case of the second, why $\mathrm{sin}{y}^{\prime}+3y+x+5=0$ isnt a first order differential equation?

asked 2022-06-22

Solve the differential equation by the appropriate substitution

${x}^{2}\frac{dy}{dx}+2xy={x}^{4}{y}^{2}+1$

${x}^{2}\frac{dy}{dx}+2xy={x}^{4}{y}^{2}+1$

asked 2022-06-04

I am trying to evaluate the equation:

${y}^{\prime}=y({y}^{2}-\frac{1}{2})$

I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?

${y}^{\prime}=y({y}^{2}-\frac{1}{2})$

I multiplied the y over and tried to solve it in seperable form (M and N). The partial deritives did not work out to be equal to eachother so I am now stuck finding an integrating factor. Is this the right approach?