What is the general solution of the differential equation $y\prime \prime +4y=0$?

kadirsmr9d
2022-09-12
Answered

What is the general solution of the differential equation $y\prime \prime +4y=0$?

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asked 2022-05-27

I was trying to compute the solution for the following differential equation:

$x(2{x}^{2}ylog(y)+1){y}^{\prime}=2y$

As I couldn't get anywhere I checked the hints in the textbook which are the following:

Reverse the way of thinking, namely view $x$ as a function and $y$ as a variable, considering that

$y=\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=>{y}^{\prime}=\frac{1}{{x}^{\prime}}$. Then it goes to say that the equation now becomes

${x}^{\prime}-\frac{x}{2y}=log(y){x}^{3}$

This final equation is obviously simple enough to solve, but how on Earth did they arrive there?

$x(2{x}^{2}ylog(y)+1){y}^{\prime}=2y$

As I couldn't get anywhere I checked the hints in the textbook which are the following:

Reverse the way of thinking, namely view $x$ as a function and $y$ as a variable, considering that

$y=\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=>{y}^{\prime}=\frac{1}{{x}^{\prime}}$. Then it goes to say that the equation now becomes

${x}^{\prime}-\frac{x}{2y}=log(y){x}^{3}$

This final equation is obviously simple enough to solve, but how on Earth did they arrive there?

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

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How to solve the differential equation $(dy/dx{)}^{2}=(x-y{)}^{2}$ with initial condition $y(0)=0$?

I solved the equation by partitioning it into two differential equations.

1) $dy/dx=x-y$ The solution is $\to $ $1-x+y=-\mathrm{exp}(-x)$

and

2) $1+x-y=\mathrm{exp}(x)$

How do we write combined solution of such equations.

I solved the equation by partitioning it into two differential equations.

1) $dy/dx=x-y$ The solution is $\to $ $1-x+y=-\mathrm{exp}(-x)$

and

2) $1+x-y=\mathrm{exp}(x)$

How do we write combined solution of such equations.