I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation:

$${u}^{\prime}+u=0\text{},\text{}y(x)=C{e}^{-x}$$

How do I tackle this?

lunja55
2022-09-22
Answered

Getting into differential equations

I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation:

$${u}^{\prime}+u=0\text{},\text{}y(x)=C{e}^{-x}$$

How do I tackle this?

I'm just getting into differential equations now and I've got to show that the given y(x) is a solution to the differential equation:

$${u}^{\prime}+u=0\text{},\text{}y(x)=C{e}^{-x}$$

How do I tackle this?

You can still ask an expert for help

Jade Mejia

Answered 2022-09-23
Author has **8** answers

Step 1

You know that the derivative of y(x)expx is a constant, so just compute it:

$$0=\frac{d}{dx}[\mathrm{exp}(x)y(x)]=\mathrm{exp}(x)y(x)+\mathrm{exp}(x){y}^{\prime}(x)$$

and as the exp is never $=0$, $y(x)+{y}^{\prime}(x)=0$.

Step 2

It also works the other way: if you consider a solution of the equation, then

$$\frac{d}{dx}[\mathrm{exp}(x)y(x)]=0$$

and then it has the given form.

You know that the derivative of y(x)expx is a constant, so just compute it:

$$0=\frac{d}{dx}[\mathrm{exp}(x)y(x)]=\mathrm{exp}(x)y(x)+\mathrm{exp}(x){y}^{\prime}(x)$$

and as the exp is never $=0$, $y(x)+{y}^{\prime}(x)=0$.

Step 2

It also works the other way: if you consider a solution of the equation, then

$$\frac{d}{dx}[\mathrm{exp}(x)y(x)]=0$$

and then it has the given form.

Corbin Bradford

Answered 2022-09-24
Author has **1** answers

Step 1

The point is that you plug $y(x)=C{e}^{-x}$ into the differential equation at the place of u , and of course the derivative of y(x) at the place of u′ (do you know how to differentiate y(x)?).

Step 2

Then you must show that the equality in the differential equation holds, that is ${y}^{\prime}(x)+y(x)=0$. So

$$\begin{array}{r}{y}^{\prime}(x)+y(x)=-C{e}^{-x}+C{e}^{-x}=(-C+C){e}^{-x}=0\cdot {e}^{-x}=0.\end{array}$$

So it is, in fact, a solution since it equals zero as it should.

The point is that you plug $y(x)=C{e}^{-x}$ into the differential equation at the place of u , and of course the derivative of y(x) at the place of u′ (do you know how to differentiate y(x)?).

Step 2

Then you must show that the equality in the differential equation holds, that is ${y}^{\prime}(x)+y(x)=0$. So

$$\begin{array}{r}{y}^{\prime}(x)+y(x)=-C{e}^{-x}+C{e}^{-x}=(-C+C){e}^{-x}=0\cdot {e}^{-x}=0.\end{array}$$

So it is, in fact, a solution since it equals zero as it should.

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