Author Topic: differential equations (Read 2680 times) Tweet Share

M-D · « **on:** June 13, 2013, 11:08:00 pm »

the questions says that $y=ax^3+bx^2+cx+d$ is a solution to $\frac{d^2y}{dx^2}+2\frac{dy}{dx} + y=x^3$ , find the constants a, b, c and d.

this is what i have done so far:

$\frac{dy}{dx}=3ax^2+2bx+c$

$\frac{d^2y}{dx^2}=6ax+2b$

if i substitute the first and second derivatives and y into $\frac{d^2y}{dx^2}+2\frac{dy}{dx} + y=x^3$ i get something which i find difficult to simplify, how should i go about this question? thanks

lzxnl · « **Reply #1 on:** June 13, 2013, 11:17:30 pm »

You sub it in, and then compare coefficients of x of both sides. See where that gets you. It may be messy though.
Remember, the equation holds for ALL values of x. You can sub any value of x you want.

b^3 · « **Reply #2 on:** June 13, 2013, 11:19:05 pm »

It's not that bad to simplify, it's just collecting like terms, and then equating the coefficients. Then the equations you get out of it, lead into one another, allowing you to solve it without much effort.

Spoiler

$\begin{alignedat}{1}y & =ax^{3}+bx^{2}+cx+d \\ \frac{dy}{dx} & =3ax^{2}+2bx+c \\ \frac{d^{2}y}{dx^{2}} & =6ax+2b \\ \frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+y & =x^{3} \\ x^{3} & =6ax+2b+2\left(3ax^{2}+2bx+c\right)+ax^{3}+bx^{2}+cx+d \\ x^{3} & =ax^{3}+\left(6a+b\right)x^{2}+\left(6a+4b+c\right)x+\left(2b+2c+d\right) \\ \text{equating coefficients} \\ a=1 \\ 6a+b & =0 \\ b & =-6a \\ \implies b & =-6 \\ 6a+4b+c & =0 \\ c & =-4b-6a \\ & =-4\left(-6\right)-6 \\ & =24-6 \\ \implies c & =18 \\ 2b+2c+d & =0 \\ d & =-2b-2c \\ & =-2\left(-6+18\right) \\ & =-24 \\ \therefore a=1,\: b=-6,\: c=18,\: d=-24 \end{alignedat}$

TrueTears · « **Reply #3 on:** June 13, 2013, 11:20:57 pm »

Also just a note to the OP, what you solve is a solution not the general solution.

lzxnl · « **Reply #4 on:** June 14, 2013, 07:39:22 pm »

Yeah...the general solution requires you to solve the same differential equation except with 0 on the RHS.

TrueTears · « **Reply #5 on:** June 14, 2013, 07:49:07 pm »

^ no?

lzxnl · « **Reply #6 on:** June 14, 2013, 08:07:58 pm »

For the general solution, don't we have to solve the homogeneous equation y''+2y'+y=0, find the general to this and then add this to the polynomial solution to y''+2y'+y=x^3?

Phy124 · « **Reply #7 on:** June 14, 2013, 08:33:20 pm »

Quote from: nliu1995 on June 14, 2013, 08:07:58 pm

For the general solution, don't we have to solve the homogeneous equation y''+2y'+y=0, find the general to this and then add this to the polynomial solution to y''+2y'+y=x^3?

Yes, as far as I know that's correct. The general solution is the sum of the complementary solution and the particular solution from what I can recall.

I believe he thought you were trying to say the general solution would be given by solving $y''(x)+2y'(x)+y(x)=0$ only.

TrueTears · « **Reply #8 on:** June 14, 2013, 09:13:59 pm »

ye you have to add it on, your sentence implied solving it standalone lol

lzxnl · « **Reply #9 on:** June 14, 2013, 10:19:57 pm »

Ah ok. Apologies for the confusion. I really meant that it was a step that was needed.

M-D · « **Reply #10 on:** June 15, 2013, 07:17:06 pm »

I've got another separable differential equation which i am finding quiet difficult to solve. Here it is:

$(xy^2-2xy)\frac{dy}{dx}=x^5y-2yxsin(2x)$ and $y=2 when x=\pi$

thanks.

BubbleWrapMan · « **Reply #11 on:** June 15, 2013, 07:29:40 pm »

Did you get as far as $\left(y-2\right)\frac{\mathrm{d}y}{\mathrm{d}x}=x^4-2\sin(2x)$ ?

M-D · « **Reply #12 on:** June 15, 2013, 08:47:39 pm »

thanks Timmeh. No i had not reached $\left(y-2\right)\frac{\mathrm{d}y}{\mathrm{d}x}=x^4-2\sin(2x)$ but i understand how you get there. i have solved the differential equation using the separation technique. this is what i got:

$y^2-4y=\frac{2}{5}x^5+2cos(2x)+c$ (i hope that is correct) now i believe i should just substitute $y=2 when x=\pi$ . is that correct? thanks

M-D · « **Reply #13 on:** June 15, 2013, 08:53:06 pm »

thanks Timmeh. I substituted y=2 when x=pi and got the right answer. thanks again.

M-D · « **Reply #14 on:** June 16, 2013, 12:06:39 pm »

sorry i got stuck with another question like my previous one. here it is:

$\frac{3}{x^2}\frac{dy}{dx}=y-\frac{3}{y^2}$ ,y does not equal 0

solve for $y=3^\frac{1}{3} when x=2$

this is what I've done thus far:

$\int\frac{y^2}{y^3-3}dy=\int\frac{x^2}{3} dx$

let $u=y^3-3$

$dy=\frac{du}{3y^2}$

therefore $\frac{1}{3}\int\frac{1}{u}du=\frac{x^3}{9} + C$

$\frac{1}{3}log(y^3-3)=\frac{x^3}{9} + C$

if I let $y=3^\frac{1}{3} when x=2$ the log part becomes undefined. what should i do?

thank you very much for your assistance.

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