guess Group Title dy/dx+y=x^2+2x solve it by 2 method one year ago one year ago

1. zepdrix

$\large y'+y=x^2+2x$ Let's start by using the method of Undetermined Coefficients. Our complimentary solution $$\large y_c$$ is given by the solution to this equation.$\large y'+y=0$ The characteristic equation is,$\large r+1=0 \qquad \rightarrow \qquad r=-1$Giving us a complimentary solution of, $$\large \qquad y_c=Ce^{-x}$$ ---------------------------------------------------------- We will also need to find the particular solution $$\large y_p$$. Our particular solution will be of the same form as the right side of our problem. In this case, we have a polynomial of degree 2. So our particular solution will look something like,$\large y_p=Ax^2+B(2x)+C$Where A, B and C are unknown constants. We'll take the derivative of our particular solution and then we can plug in $$\large y_p$$ and $$\large y_p^{'}$$ into the original problem to solve for A, B and C.

2. zepdrix

Eventually we're trying to get an answer that consists of the complimentary AND particular solution. $$\large \qquad y=y_c+y_p$$

3. zepdrix

$\large y_p=Ax^2+2Bx+C$$\large y_p^{'}=2Ax+2B$

4. zepdrix

Plugging these into the original equation gives us,$y'+y=x^2+2x \qquad \rightarrow \qquad \left(2Ax+2B\right)+\left(Ax^2+2Bx+C\right)=x^2+2x$

5. zepdrix

From here we can equate like terms.$2B+C=0$$2Ax+2Bx=2x \qquad \rightarrow \qquad 2A+2B=2$$Ax^2=x^2 \qquad \rightarrow \qquad A=1$

6. zepdrix

Does any of this make sense @guess or should i stop? XD

7. guess

@zepdrix thank you ,but if you can solve it by laplace Be very grateful

8. guess

@zepdrix yes the particular solution i think that what he want thanks and don't stop please:)

9. nitz

u need any help........

10. guess

11. hartnn

u know whats L[dy/dx] =... ?

12. hartnn

assuming initial conditions =0 L[dy/dx] = s Y(s) where Y(s) is the Laplace Transform of y(t). so, take Laplace on both sides of dy/dx+y=x^2+2x and can you tell me what you get in first step ??

13. guess

@zepdrix then what ? is b=1/2 ?

14. guess

@zepdrix @hartnn @nitz this question came at midterm and ididn't solve it ): help me please

15. Callisto

Can't we use integrating factor to solve it?

16. guess

@Callisto mm Can you show me?

17. mukushla

what Callisto said probabely is the best bet :)

18. Callisto

$y'+p(x)y=q(x)$ $\alpha = exp[\int p(x)dx]$$y=\frac{1}{\alpha}\int \alpha [q(x)] dx$

19. Callisto

in your case p(x)=1, q(x) = x^2+2x

20. guess

@Callisto it look like will be best bet really !! continue please :)

21. Callisto

Ha! Several integration by parts there :S $y'+y=x^2+2x$ $\alpha = e^{\int 1 dx}=e^x$ $y=e^{-x}\int e^x(x^2+2x)dx$$=e^{-x}[e^x(x^2+2x)-\int e^x(2x+2)dx]$$=e^{-x}[e^x(x^2+2x)-2(e^x(x+1)-\int e^xdx)]$$=...$

22. Callisto

@guess Can you do it from here? Or still need more help?

23. hartnn

using Laplace $$L[dy/dx]+L[y]=L[x^2]+2L[x] \\sY+Y=2/s^3+2/s \\ Y=2(1+s^2)/[s^3(s+1)] \\ Y=2/x^3-2/x^2+4/x-4/(s+1)\\\text{skipping the partial fractions part for you.Taking Inverse Laplace}\\ y = x^2-2x+4-4e^x$$ Someone please verify the final answer @zepdrix or @Callisto

24. hartnn

$$y=e^{-x}\int e^x(x^2+2x)dx , after \:\: simplification, y=x^2$$

25. guess

@hartnn @Callisto yes please continue i got this put i feel it Error exp^(- x) {x^2e^x-2e^x} and thank alot for all

26. Callisto

Method of integrating factor: y'+y=x^2+2x $\alpha =e^x$$y=e^{-x}\int e^x(x^2+2x)dx$$=e^{-x}[e^x(x^2+2x)-2\int e^x(x+1)]$$=e^{-x}[e^x(x^2+2x)-2e^x(x+1)+2e^x+C]$$=e^{-x}(e^xx^2+C)$$=x^2+Ce^{-x}$

27. guess

thank you so much @Callisto @hartnn you're very helper :))

28. guess

by laplace L[ y`]+L[y]=L[x2]+2L[x] S(Y)+Y(S)=2/S^3+2/S^2 (S+1)Y(S)=2(S+1)/S^3 y(S)=2/S^3 y(x)=x^2 @hartnn @Callisto right ??

29. hartnn

ohh.....yes!