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zepdrix Group TitleBest ResponseYou've already chosen the best response.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, \(\large \qquad y_c=Ce^{x}\)  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.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Eventually we're trying to get an answer that consists of the complimentary AND particular solution. \(\large \qquad y=y_c+y_p\)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large y_p=Ax^2+2Bx+C\]\[\large y_p^{'}=2Ax+2B\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Plugging these into the original equation gives us,\[y'+y=x^2+2x \qquad \rightarrow \qquad \left(2Ax+2B\right)+\left(Ax^2+2Bx+C\right)=x^2+2x\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
From here we can equate like terms.\[2B+C=0\]\[2Ax+2Bx=2x \qquad \rightarrow \qquad 2A+2B=2\]\[Ax^2=x^2 \qquad \rightarrow \qquad A=1\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Does any of this make sense @guess or should i stop? XD
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@zepdrix thank you ,but if you can solve it by laplace Be very grateful
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@zepdrix yes the particular solution i think that what he want thanks and don't stop please:)
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
u need any help........
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@nitz yes please
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
u know whats L[dy/dx] =... ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
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 ??
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@zepdrix then what ? is b=1/2 ?
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@zepdrix @hartnn @nitz this question came at midterm and ididn't solve it ): help me please
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Can't we use integrating factor to solve it?
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@Callisto mm Can you show me?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
what Callisto said probabely is the best bet :)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
\[y'+p(x)y=q(x)\] \[\alpha = exp[\int p(x)dx]\]\[y=\frac{1}{\alpha}\int \alpha [q(x)] dx\]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
in your case p(x)=1, q(x) = x^2+2x
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@Callisto it look like will be best bet really !! continue please :)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Ha! Several integration by parts there :S \[y'+y=x^2+2x\] \[\alpha = e^{\int 1 dx}=e^x\] \[y=e^{x}\int e^x(x^2+2x)dx\]\[=e^{x}[e^x(x^2+2x)\int e^x(2x+2)dx]\]\[=e^{x}[e^x(x^2+2x)2(e^x(x+1)\int e^xdx)]\]\[=...\]
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
@guess Can you do it from here? Or still need more help?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
using Laplace \(L[dy/dx]+L[y]=L[x^2]+2L[x] \\sY+Y=2/s^3+2/s \\ Y=2(1+s^2)/[s^3(s+1)] \\ Y=2/x^32/x^2+4/x4/(s+1)\\\text{skipping the partial fractions part for you.Taking Inverse Laplace}\\ y = x^22x+44e^x\) Someone please verify the final answer @zepdrix or @Callisto
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
\(y=e^{x}\int e^x(x^2+2x)dx , after \:\: simplification, y=x^2\)
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
@hartnn @Callisto yes please continue i got this put i feel it Error exp^( x) {x^2e^x2e^x} and thank alot for all
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.4
Method of integrating factor: y'+y=x^2+2x \[\alpha =e^x\]\[y=e^{x}\int e^x(x^2+2x)dx\]\[=e^{x}[e^x(x^2+2x)2\int e^x(x+1)]\]\[=e^{x}[e^x(x^2+2x)2e^x(x+1)+2e^x+C]\]\[=e^{x}(e^xx^2+C)\]\[=x^2+Ce^{x}\]
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
thank you so much @Callisto @hartnn you're very helper :))
 one year ago

guess Group TitleBest ResponseYou've already chosen the best response.1
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 ??
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
ohh.....yes!
 one year ago
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