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zepdrix
 2 years ago
Best 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.

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

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

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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Does any of this make sense @guess or should i stop? XD

guess
 2 years ago
Best ResponseYou've already chosen the best response.1@zepdrix thank you ,but if you can solve it by laplace Be very grateful

guess
 2 years ago
Best 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:)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1u know whats L[dy/dx] =... ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1assuming 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 ??

guess
 2 years ago
Best ResponseYou've already chosen the best response.1@zepdrix then what ? is b=1/2 ?

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

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.4Can't we use integrating factor to solve it?

guess
 2 years ago
Best ResponseYou've already chosen the best response.1@Callisto mm Can you show me?

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0what Callisto said probabely is the best bet :)

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

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

guess
 2 years ago
Best ResponseYou've already chosen the best response.1@Callisto it look like will be best bet really !! continue please :)

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

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.4@guess Can you do it from here? Or still need more help?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1using 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

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

guess
 2 years ago
Best 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

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

guess
 2 years ago
Best ResponseYou've already chosen the best response.1thank you so much @Callisto @hartnn you're very helper :))

guess
 2 years ago
Best ResponseYou've already chosen the best response.1by 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 ??
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