What is the general solution of the differential equation $\frac{dy}{dx}-2y+a=0$?

tamolam8
2022-09-12
Answered

What is the general solution of the differential equation $\frac{dy}{dx}-2y+a=0$?

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Raphael Singleton

Answered 2022-09-13
Author has **19** answers

First write the DE in standard form:

$\frac{dy}{dx}-2y+a=0$

$\therefore \frac{dy}{dx}-2y=-a$......[1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

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

$={e}^{\int -2dx}$

$={e}^{-2x}$

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

$\frac{dy}{dx}{e}^{-2x}-2y{e}^{-2x}=-a{e}^{-2x}$

$\frac{d}{dx}\left(y{e}^{-2x}\right)=-a{e}^{-2x}$

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

$y{e}^{-2x}=\int -a{e}^{-2x}dx$

Which we can easily integrate to get:

$y{e}^{-2x}=\frac{1}{2}a{e}^{-2x}+C$

$\therefore y=\frac{1}{2}a+C{e}^{2x}$

$\frac{dy}{dx}-2y+a=0$

$\therefore \frac{dy}{dx}-2y=-a$......[1]

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

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

This is a standard form of a Differential Equation that can be solved by using an Integrating Factor:

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

$={e}^{\int -2dx}$

$={e}^{-2x}$

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

$\frac{dy}{dx}{e}^{-2x}-2y{e}^{-2x}=-a{e}^{-2x}$

$\frac{d}{dx}\left(y{e}^{-2x}\right)=-a{e}^{-2x}$

This has converted our DE into a First Order separable DE which we can now just separate the variables to get;

$y{e}^{-2x}=\int -a{e}^{-2x}dx$

Which we can easily integrate to get:

$y{e}^{-2x}=\frac{1}{2}a{e}^{-2x}+C$

$\therefore y=\frac{1}{2}a+C{e}^{2x}$

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How do you find the general solution to $\frac{dy}{dx}+{e}^{x+y}=0$?

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Let the first order differential equation be

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

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Find the general solution to the equation

$${x}^{2}{y}^{\prime}+3xy=1$$

$${x}^{2}{y}^{\prime}+3xy=1$$

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What is the general solution of the differential equation $\frac{dy}{dx}+2y=0$?

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I have this simple differential equation: ${y}^{\prime}=(\mathrm{tan}x)y.$

after integrating $\frac{{y}^{\prime}}{y(x)}=\mathrm{tan}x$

i came up with $\mathrm{log}y(x)=-\mathrm{log}\mathrm{cos}(x)+1.$

now my question is this one: why ${e}^{-\mathrm{log}\mathrm{cos}(x)}=\mathrm{sec}(x)?$

after integrating $\frac{{y}^{\prime}}{y(x)}=\mathrm{tan}x$

i came up with $\mathrm{log}y(x)=-\mathrm{log}\mathrm{cos}(x)+1.$

now my question is this one: why ${e}^{-\mathrm{log}\mathrm{cos}(x)}=\mathrm{sec}(x)?$