Differential equation and initial value

[-EquinoX-]

Homework Statement

[img=http://img3.imageshack.us/img3/3417/questionec.th.jpg]

Homework Equations

The Attempt at a Solution

I tried dividing both sides by (x^5+1), however the integration becomes really complex... can someone give me suggestions on how to do this?

 

[rock.freak667]

I tried dividing both sides by (x^5+1), however the integration becomes really complex... can someone give me suggestions on how to do this?

put the -10x⁴y on the left side then divide by tr x⁵+1. Integrating factor it.

 

[-EquinoX-]

that's what I did and then I need to integrate -10x^4/(x^5+1) dx right and then then to get the integrating factor is just e to the power of whatever the result of the integration is... however the integration is quite hard...

 

[rock.freak667]

that's what I did and then I need to integrate -10x^4/(x^5+1) dx right and then then to get the integrating factor is just e to the power of whatever the result of the integration is... however the integration is quite hard...

alright now, so


$$\int \frac{-10x^4}{x^5+1} dx$$

see how d/dx(x⁵+1)=5x⁴ ?

Can you use a substitution to make this integral easier?

 

[-EquinoX-]

alright now, so


$$\int \frac{-10x^4}{x^5+1} dx$$

see how d/dx(x⁵+1)=5x⁴ ?

Can you use a substitution to make this integral easier?

Okay say I solve the integral and then the integrating factor would be e to the power of this resulting integral right? so then what do I need to do next in order to solve for this problem?

 

[rock.freak667]

Okay say I solve the integral and then the integrating factor would be e to the power of this resulting integral right? so then what do I need to do next in order to solve for this problem?

right so for y'+P(x)y=Q(x), when you multiply by an integrating factor 'u', the left side becomes d/dx(uy) . That's why we multiply by u in the first place.


So you'll need to basically integrate uQ(x) w.r.t. x

 

[-EquinoX-]

and then divide that by the integrating factor right?

 

[rock.freak667]

and then divide that by the integrating factor right?

yes you can if you feel the need to.

 

[-EquinoX-]

what do you mean by if I need to? doesn't it always works like that?

 

[-EquinoX-]

this now comes to:

$$\int \frac{x^2+5x-4}{x^5+1} e^{-2ln(x^5+1)} dx$$

I guess this can be simplified to:

$$\int \frac{x^2+5x-4}{(x^5+1)^3} dx$$

is this true?

how can I solve this such complex integration?

 

[-EquinoX-]

anyone please?

 

[rock.freak667]

anyone please?

Re do your integrating factor as it was to integrate 10x⁴/(x⁵+1) not with the -ve sign.

 

[-EquinoX-]

Integrating $$\frac{10x^4}{(x^5+1)}$$ the result I got is $$2ln(x^5+1)$$ and so the integrating factor is $$e^{2ln(x^5+1)}$$ which simplifies to $$(x^5+1)^2$$.

Then I do $$\int (x^2+5x-4)(x^5+1)$$ and the result of this integration I divide by $$(x^5+1)^2$$ which is the integrating factor. Is this the correct step to find the solution?

Please correct me if I am wrong.

 

[rock.freak667]

Integrating $$\frac{10x^4}{(x^5+1)}$$ the result I got is $$2ln(x^5+1)$$ and so the integrating factor is $$e^{2ln(x^5+1)}$$ which simplifies to $$(x^5+1)^2$$.

Then I do $$\int (x^2+5x-4)(x^5+1)$$ and the result of this integration I divide by $$(x^5+5)^2$$ which is the integrating factor. Is this the correct step to find the solution?

Please correct me if I am wrong.

yes just integrate and divide, also don't forget the constant of integration

 

[-EquinoX-]

yes just integrate and divide, also don't forget the constant of integration

by the constant of integration you mean the C right? The terms after integration that I found is very long...