General solution of this equation

[c4l3b]

I am having problems trying to find general solution of this equation;

\(\displaystyle \frac{dx}{dt}[\tex]= e^t/2 + 4te^2t^2+3 + t^5.

Find the particular solution which satisfies x(0) = 16


I have differentiated e^t/2 but the rest I am really confused :?

Could anybody show me the calculations?

Thanks alot!\)

 

[Deleted member 4993]

I am having problems trying to find general solution of this equation;

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} + 4te^2t^2+3 + t^5\) ...............Is this what your problem looks like?

Find the particular solution which satisfies x(0) = 16


I have differentiated e^t/2 but the rest I am really confused :?

Could anybody show me the calculations?

Thanks alot!

 

[galactus]

You need to integrate, not differentiate.

 

[c4l3b]

I am having problems trying to find general solution of this equation;

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} + 4te^2t^2+3 + t^5\) ...............Is this what your problem looks like?

Find the particular solution which satisfies x(0) = 16


I have differentiated e^t/2 but the rest I am really confused :?

Could anybody show me the calculations?

Thanks alot!

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} +4te^2^t^2^+^3+t^5\)

2t2 + 3 is meant to be 2t squared then + 3 inline with 2t....I did not know how to implement this in latex form

 

[Deleted member 4993]

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} +4te^2^t^2^+^3+t^5\)

2t2 + 3 is meant to be 2t squared then + 3 inline with 2t....I did not know how to implement this in latex form

So it is:

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} + 4te^{2t^2+3} + t^5\)

Now intigrate part by part:

\(\displaystyle \int e^{t/2} \, dt \, = \, ???\)

and

\(\displaystyle \int 4te^{2t^2+3} \, dt \, = \, ???\)

and so on....

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[c4l3b]

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} +4te^2^t^2^+^3+t^5\)

2t2 + 3 is meant to be 2t squared then + 3 inline with 2t....I did not know how to implement this in latex form

So it is:

\(\displaystyle \frac{dx}{dt} \, = \, e^{t/2} + 4te^{2t^2+3} + t^5\)

Now intigrate part by part:

\(\displaystyle \int e^{t/2} \, dt \, = \, ???\)

and

\(\displaystyle \int 4te^{2t^2+3} \, dt \, = \, ???\)

and so on....

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

Ok;

so for \(\displaystyle \int e^{t/2} \, dt \, = \, ???\)

I factored out constants; e^t/2+constant

and \(\displaystyle \int 4te^{2t^2+3} \, dt \, = \, ???\) = struggled with this one could anyone show me the working out?


\(\displaystyle \int t^5\) = t^6/6 +constant

 

[Deleted member 4993]

Ok;

so for \(\displaystyle \int e^{t/2} \, dt \, = \, ???\)

I factored out constants; e^t/2+constant

Incorrect

If you differentiate e^(t/2) + C ? you get (1/2)* e^(t/2) - not the original e^(t/2).

It should be 2 * e^(t/2) + C

and \(\displaystyle \int 4te^{2t^2+3} \, dt \, = \, ???\) = struggled with this one could anyone show me the working out?

substitute

u = 2t^2 + 3

du = 4t dt, then

\(\displaystyle \int 4te^{2t^2+3} \, dt \, = \, \int e^{2t^2+3} \, (4t dt) \, = \, \int e^{u} \, du \, = e^u \, + \, C \, = e^{2t^2+3} \, + \, C\)

\(\displaystyle \int t^5\) = t^6/6 +constant <<< Correct

Now put those all together... and apply the initial condition x(0) = 16 to solve for constant