Differental Equation Question, Particular Solutions?

[stripedcat]

\(\displaystyle dy/dx = x^2+x^2y, y(1)=2\)

I'm not sure what to do with the y(1)=2, presumably I want to do the equation and then find y and c?

\(\displaystyle dy/dx=x^2(y+1)\)

\(\displaystyle dy/y+1=x^2 dx\)

\(\displaystyle ln(y+1)=x^3/3+C\)

\(\displaystyle y=e^{x^3/3} - 1 + C\)

I'm assuming I did this part right?

I just plug in 2 where the y is and I where the x is, then solve for y and c? Is that how it works?

 

[Saitama]

Re: Differiental Equation Question, Particular Solutions?

\(\displaystyle dy/dx = x^2+x^2y, y(1)=2\)

I'm not sure what to do with the y(1)=2, presumably I want to do the equation and then find y and c?

\(\displaystyle dy/dx=x^2(y+1)\)

\(\displaystyle dy/y+1=x^2 dx\)

\(\displaystyle ln(y+1)=x^3/3+C\)

Correct so far.

\(\displaystyle y=e^{x^3/3} - 1 + C\)

This should be $y=Ae^{x^3/3}-1$ where $A$ is some constant.

Now just use the initial condition to determine $A$.

 

[stripedcat]

Re: Differiental Equation Question, Particular Solutions?

Correct so far.

This should be $y=Ae^{x^3/3}-1$ where $A$ is some constant.

Now just use the initial condition to determine $A$.

So like this...?

\(\displaystyle 2=Ce^{1^3/3}-1\)

Solve for C

\(\displaystyle e^{1/3} = 1.3956\)

\(\displaystyle 2=C(1.3956)-1\)

\(\displaystyle 3=C(1.39561)\)

C=~2.1496

Now that I have C...

\(\displaystyle y=(2.1496)(1.3956)-1\)

y=-0.24601?

 

[Saitama]

Re: Differiental Equation Question, Particular Solutions?

So like this...?

\(\displaystyle 2=Ce^{1^3/3}\)

Shouldn't that be $3=Ce^{1/3}$?

 

[stripedcat]

Re: Differiental Equation Question, Particular Solutions?

Shouldn't that be $3=Ce^{1/3}$?

Forgot the -1

But beyond that did I plug in the values like I was suppose to?

 

[Saitama]

Re: Differiental Equation Question, Particular Solutions?

Now that I have C...

\(\displaystyle y=(2.1496)(1.3956)-1\)

y=-0.24601?

Any reason to replace $e^{x^3/3}$ with $1.3956$? You got $C$ in a correct way, so the final answer is $y=1.3956e^{x^3/3}-1$, do you see why?

 

[stripedcat]

Re: Differiental Equation Question, Particular Solutions?

So I got C, that's good then! Thank you!

Do we not plug in the previous X value when trying to solve for Y? I was wondering about that, it didn't seem right, but then by not including it I didn't see how to solve for Y without it being an abstract answer rather than a specific one.

So it's just the \(\displaystyle y=e^{x^3/3} -1\) for y then. Okay.

 

[Saitama]

Re: Differiental Equation Question, Particular Solutions?

Do we not plug in the previous X value when trying to solve for Y? I was wondering about that, it didn't seem right, but then by not including it I didn't see how to solve for Y without it being an abstract answer rather than a specific one.

I am not sure what you ask here. The condition $y(1)=2$ was given to determine $A$. $y(1)=2$ means that at $x=1$, $y=2$, so you plug in these values of $x$ and $y$ and find $A$.

So it's just the \(\displaystyle y=e^{x^3/3} -1\) for y then. Okay.

No, look at my previous post.