can someone please help me solve this question

[tribal epoch]

A, B, and D require a change of variable. C is already separable.

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I'm still not good at this subject can you please solve them on a paper if you can🙏

[tribal epoch]

!noans

[native orchid]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[thick current]

I know but Idk how can I solve it

[tribal epoch]

Do you know how to solve separable differential equations?

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Unfortunately no

I haven't studied that yet

[tribal epoch]

How are you getting these problems then?

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Self Learning mostly and I asked my teacher to give me some questions (it's not a homework or like that)

[tribal epoch]

Do you know integration at least?

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I know Bernoulli, Riccati, Claro, Lagrange these type of questions

[tribal epoch]

Okay. So something simple like $\int \cos x dx$, you can solve that?

[crystal ridge]

SWR

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Of course that's sin(x) + C

[tribal epoch]

perfect

okay then I'll give you quick info a simple differential equation, then a seperable differential equation.

Assume we have the equation $y'=f(x)$ and we wanted to solve for $y$. How we can do this is by integrating both sides. If $y'=f(x)$, then $\int y' dy=\int f(x)dx+C$ where $C$ is any constant.

[crystal ridge]

SWR

[tribal epoch]

And, of course, $\int y'dy=y$ by definition of derivative of and antiderivative

[crystal ridge]

SWR

[tribal epoch]

Now here's the neat part. Assume instead of $y'=f(x)$, we wanted to solve $y'=f(x, y)$. Now, $f$ is a function of both $x$ and $y$.

[crystal ridge]

SWR

[tribal epoch]

In general, we cannot always solve this. But, assume we could write $f(x, y)$ as a product of a function of $x$ and a function of $y$. That is, assume there are functions $g(x)$ and $h(y)$ such that $f(x, y)=g(x)h(y)$.

[crystal ridge]

SWR

[tribal epoch]

Then $f$ is called a separable function, and $y'=g(x)h(y)$ is separable differentiable equation.

[crystal ridge]

SWR

[tribal epoch]

This type of equation is actually pretty easy to solve. Just divide both sides by $h(y)$ (where we assume $h(y)\ne 0$). Then we have $\frac{y'}{h(y)}=g(x)$

[crystal ridge]

SWR

[tribal epoch]

I need to make a minor correction here. Instead of $y'$, we'll write $\frac{dy}{dx}$. Then we can write $\frac{dy}{dx}=f(x)$ as $dy=f(x)dx$. We then integrate both sides of this to get $\int dy=\int f(x)dx+C$, or $y=\int f(x)dx+C$

[crystal ridge]

SWR

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I see

[tribal epoch]

With that correction, I'll rewrite this separable equation as $\frac{dy}{h(y)}=g(x)dx$. Now the left side is just in terms of $y$, and the right side is just in terms of $x$ (this is why we call the differential equation "separable"). We can solve this now just by integrating both sides: $\int{\frac{dy}{h(y)}}=\int f(x)dx+C$

[crystal ridge]

SWR

[tribal epoch]

As a simple exercise of your understanding, try solving $y'=xy$.

[crystal ridge]

SWR

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Can we say ln(y) = $x*dx ==> ln(y) = (1/2)*x^2 and then *e to both sides => y = e^((1/2)*x^2)

[tribal epoch]

close.

You need the +C

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Oh,right I forgot sorry

Thank you 🙏

[tribal epoch]

It's more than $+C$ just fyi. This part is extremely important.

[crystal ridge]

SWR

[tribal epoch]

We have $\frac{dy}{y}=xdx$. We integrate both sides and $\ln\abs{y}=\frac{1}{2}x^2+C$

[crystal ridge]

SWR

[tribal epoch]

We now exponentiate both sides: $\abs{y}=\exp(\frac{1}{2}x^2+C)$

[crystal ridge]

SWR

[thick current]

Thanks alot

[tribal epoch]

See how the +C is inside the exponent?

[thick current]

Yes

[tribal epoch]

Yeah. That's pretty vital. It changes the equation greatly.

You can distribute the exponent to get $\abs{y}=e^C e^{x^2/2}$

[crystal ridge]

SWR

[thick current]

Yeah

[tribal epoch]

Then you can remove the absolute value by instead writing $y=\pm e^C e^{x^2/2}$

[crystal ridge]

SWR

[tribal epoch]

But $\pm e^C$ is just a constant value based on $C$, so you can write the whole thing as its own constant: $A$ (or some people like to do lowercase $c$). So the full final solution would be $y=Ae^{x^2/2}$

sound good?

[crystal ridge]

SWR

[thick current]

I see thanks again

[tribal epoch]

So, see if you can do C. It's a separable equation. Try to get all the x and y terms separated