William Collins
2021-12-29
Answered

Solve for the differential equations and get the general solution. Simplify your answer free from In. $dy=\mathrm{tan}x\mathrm{tan}ydx$

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vrangett

Answered 2021-12-30
Author has **36** answers

Answer: The general solution is

Hattie Schaeffer

Answered 2021-12-31
Author has **37** answers

Given differential equation

$dy=\mathrm{tan}x\mathrm{tan}y\text{}dx$

$\Rightarrow \frac{dy}{\mathrm{tan}y}=\mathrm{tan}x\text{}dx$

$\Rightarrow \frac{\mathrm{cos}y}{\mathrm{sin}y}dy=\frac{\mathrm{sin}x}{\mathrm{cos}x}dx$

Integrating,$\int \frac{\mathrm{cos}y}{\mathrm{sin}y}dy=\int \frac{\mathrm{sin}x}{\mathrm{cos}x}dx$ (1)

We substitute left hand integral

$u=\mathrm{sin}y$

$\Rightarrow du=\mathrm{cos}y\text{}dy$

and right hand integral

$\mathrm{cos}x=v$

$\Rightarrow -\mathrm{sin}x\text{}dx=dv$

$\Rightarrow \mathrm{sin}x\text{}dx=-dv$

Then from (1)

$\int \frac{du}{u}=\int \frac{-dv}{v}$

$\Rightarrow \mathrm{ln}\left(u\right)=-\mathrm{ln}\left(v\right)+\mathrm{ln}\left(c\right)$

$\Rightarrow \mathrm{ln}\left(u\right)=\mathrm{ln}\left(\frac{c}{v}\right)$

$\Rightarrow u=\frac{c}{v}$

$\Rightarrow vu=c$

$\Rightarrow \mathrm{cos}x\mathrm{sin}y=c$

Hence solution is$\mathrm{cos}x\mathrm{sin}y=c$

Integrating,

We substitute left hand integral

and right hand integral

Then from (1)

Hence solution is

karton

Answered 2022-01-09
Author has **439** answers

We know,

asked 2022-09-13

Find ${\mathcal{L}}^{-1}(F(s))$, if given $F(s)={\displaystyle \frac{2}{s({s}^{2}+4)}}$

I have tried as below.

To find inverse of Laplace transform, I want to make partial fraction as below.

$$\begin{array}{r}{\displaystyle \frac{2}{s({s}^{2}+4)}}={\displaystyle \frac{A}{s}}+{\displaystyle \frac{Bs+C}{{s}^{2}+4}}={\displaystyle \frac{(A+B){s}^{2}+Cs+4A}{s({s}^{2}+4)}}.\end{array}$$

After that, we have system of linear equation

$$\begin{array}{rl}A+B& =0\\ C& =0\\ 4A& =2.\end{array}$$

Thus we have A=2, B=−2, and C=0. Now, substituting A,B,C and we have

$$\begin{array}{r}{\displaystyle \frac{2}{s({s}^{2}+4)}}={\displaystyle \frac{2}{s}}+{\displaystyle \frac{-2s}{{s}^{2}+4}}.\end{array}$$

But the fact is

$$\begin{array}{r}{\displaystyle \frac{2}{s({s}^{2}+4)}}\ne {\displaystyle \frac{2}{s}}+{\displaystyle \frac{-2s}{{s}^{2}+4}}={\displaystyle \frac{8}{{s}^{2}+4}}.\end{array}$$

I'm stuck here. I can't make a partial fraction for F(s) and I can't find inverse of Laplace transform for F(s).

Anyone can give me hint to give me hint for this problem?

I have tried as below.

To find inverse of Laplace transform, I want to make partial fraction as below.

$$\begin{array}{r}{\displaystyle \frac{2}{s({s}^{2}+4)}}={\displaystyle \frac{A}{s}}+{\displaystyle \frac{Bs+C}{{s}^{2}+4}}={\displaystyle \frac{(A+B){s}^{2}+Cs+4A}{s({s}^{2}+4)}}.\end{array}$$

After that, we have system of linear equation

$$\begin{array}{rl}A+B& =0\\ C& =0\\ 4A& =2.\end{array}$$

Thus we have A=2, B=−2, and C=0. Now, substituting A,B,C and we have

$$\begin{array}{r}{\displaystyle \frac{2}{s({s}^{2}+4)}}={\displaystyle \frac{2}{s}}+{\displaystyle \frac{-2s}{{s}^{2}+4}}.\end{array}$$

But the fact is

$$\begin{array}{r}{\displaystyle \frac{2}{s({s}^{2}+4)}}\ne {\displaystyle \frac{2}{s}}+{\displaystyle \frac{-2s}{{s}^{2}+4}}={\displaystyle \frac{8}{{s}^{2}+4}}.\end{array}$$

I'm stuck here. I can't make a partial fraction for F(s) and I can't find inverse of Laplace transform for F(s).

Anyone can give me hint to give me hint for this problem?

asked 2020-12-27

Obtain the Laplace Transform of

$L\{{e}^{-2x}+4{e}^{-3x}\}$

asked 2022-01-21

Solving Differential Equation

$(t+2)dx=2{x}^{2}dt$

asked 2021-12-17

A curve passes through the point (0, 5) and has the property
that the slope of the curve at every point P is twice the
y-coordinate of P. What is the equation of the curve?

asked 2021-12-31

Solve the differential equations $(t-1)3d\frac{s}{dt}+4(t-1)2s=t+1,t>1$

asked 2022-09-11

How to find f using Laplace transformation?

$f={J}_{0}\ast {J}_{0}$ where * is a convolution. According to the Convolution theorem it is

$$({J}_{0}\ast {J}_{0})(t):={\int}_{0}^{t}{J}_{0}(t-\tau ){J}_{0}(\tau )\mathrm{d}\tau $$

$${J}_{\nu}(z)={\left(\frac{z}{2}\right)}^{\nu}\sum _{k=0}^{\mathrm{\infty}}\frac{(-1{)}^{k}}{k!\mathrm{\Gamma}(\nu +k+1)}{\left(\frac{z}{2}\right)}^{2k}$$

$f={J}_{0}\ast {J}_{0}$ where * is a convolution. According to the Convolution theorem it is

$$({J}_{0}\ast {J}_{0})(t):={\int}_{0}^{t}{J}_{0}(t-\tau ){J}_{0}(\tau )\mathrm{d}\tau $$

$${J}_{\nu}(z)={\left(\frac{z}{2}\right)}^{\nu}\sum _{k=0}^{\mathrm{\infty}}\frac{(-1{)}^{k}}{k!\mathrm{\Gamma}(\nu +k+1)}{\left(\frac{z}{2}\right)}^{2k}$$

asked 2021-02-16

find the inverse Laplace transform of the given function.

$F\left(s\right)=\frac{2s-3}{{s}^{2}-4}$