Linearizing $y\prime \prime +\mathrm{sin}(2x+\mathrm{cos}(2y\prime +y))+1-\mathrm{sin}(y+3y\prime )=0$

Mary Buchanan
2022-01-20
Answered

Linearizing $y\prime \prime +\mathrm{sin}(2x+\mathrm{cos}(2y\prime +y))+1-\mathrm{sin}(y+3y\prime )=0$

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Tiefdruckot

Answered 2022-01-20
Author has **46** answers

The easiest linearization I can think of in this case is to use $\mathrm{cos}\left(\theta \right)\approx 1,\text{and}\text{}\mathrm{sin}\left(\theta \right)\approx \theta \text{}\text{where}\text{}\theta$ is small enough. In your case, these assumptions boil down to making the assumption that y, ${y}^{\prime}\approx 0$ .

With these assumptions, your differential equation now becomes$y{}^{\u2033}+\mathrm{sin}(2x+1)+1-(y+3{y}^{\prime})=0$

which in turn becomes

$y{}^{\u2033}-3{y}^{\prime}-y=-(1+\mathrm{sin}(2x+1))$

Now you can use your differential equation tricks to solve the above equation to get

$y\left(x\right)={c}_{1}\mathrm{exp}\left(\left(\frac{3-\sqrt{13}}{2}\right)x\right)+{c}_{2}\mathrm{exp}$

$\left(\left(\frac{3+\sqrt{13}}{2}\right)x\right)+\frac{5\mathrm{sin}(2x+1)-6\mathrm{cos}(2x+1)+61}{61}$

With these assumptions, your differential equation now becomes

which in turn becomes

Now you can use your differential equation tricks to solve the above equation to get

asked 2021-12-16

Solve the given differential equation by undetermined coefficients

$y{}^{\u2033}+7{y}^{\prime}+6y=30$

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The Laplace transform of a function f=f(x) has the following definition:

$$\begin{array}{}\text{(1)}& \mathcal{L}(f(x))={\int}_{0}^{\mathrm{\infty}}{e}^{-sx}f(x)dx\end{array}$$

However, when $f=f(ax-c)$, where a and c are arbitrary constants, we have

$$\begin{array}{}\text{(2)}& \mathcal{L}(f(ax-c))={\int}_{0}^{\mathrm{\infty}}{e}^{-sx}f(ax-c)dx\end{array}$$

with which I got very confused, why should (2) not be

$$\begin{array}{}\text{(3)}& \mathcal{L}(f(ax-c))={\int}_{0}^{\mathrm{\infty}}{e}^{-s(ax-c)}f(ax-c)d(ax-c)\end{array}$$

or equivalently

$$\begin{array}{}\text{(3)}& \mathcal{L}(f(ax-c))=a{\int}_{0}^{\mathrm{\infty}}{e}^{-s(ax-c)}f(ax-c)dx\end{array}$$

instead?

$$\begin{array}{}\text{(1)}& \mathcal{L}(f(x))={\int}_{0}^{\mathrm{\infty}}{e}^{-sx}f(x)dx\end{array}$$

However, when $f=f(ax-c)$, where a and c are arbitrary constants, we have

$$\begin{array}{}\text{(2)}& \mathcal{L}(f(ax-c))={\int}_{0}^{\mathrm{\infty}}{e}^{-sx}f(ax-c)dx\end{array}$$

with which I got very confused, why should (2) not be

$$\begin{array}{}\text{(3)}& \mathcal{L}(f(ax-c))={\int}_{0}^{\mathrm{\infty}}{e}^{-s(ax-c)}f(ax-c)d(ax-c)\end{array}$$

or equivalently

$$\begin{array}{}\text{(3)}& \mathcal{L}(f(ax-c))=a{\int}_{0}^{\mathrm{\infty}}{e}^{-s(ax-c)}f(ax-c)dx\end{array}$$

instead?

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How to prove the following identity regarding Laplace transforms?

$$\mathcal{L}[{\int}_{0}^{x}f(x-t)g(t)\text{}dt]=F(p)G(p)$$

$$\mathcal{L}[{\int}_{0}^{x}f(x-t)g(t)\text{}dt]=F(p)G(p)$$

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$f(t)=0\text{if}t2\text{or if}t3$

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The solution of the above equation is given as

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