Question

dy/dx+y=x^2+2x solve it by 2 method

Answer 1

\[\large y'+y=x^2+2x\]
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, \(\large \qquad y_c=Ce^{-x}\)
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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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

@nitz yes please

Answer 11

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

Answer 12

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 ??

Answer 13

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

Answer 14

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

Answer 15

Can't we use integrating factor to solve it?

Answer 16

@Callisto mm Can you show me?

Answer 17

what Callisto said probabely is the best bet :)

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Ha! Several integration by parts there :S

\[y'+y=x^2+2x\]
\[\alpha = e^{\int 1 dx}=e^x\]
\[y=e^{-x}\int e^x(x^2+2x)dx\]\[=e^{-x}[e^x(x^2+2x)-\int e^x(2x+2)dx]\]\[=e^{-x}[e^x(x^2+2x)-2(e^x(x+1)-\int e^xdx)]\]\[=...\]

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

@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

Answer 26

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

Answer 27

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

Answer 28

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 ??

Answer 29

ohh.....yes!