[SOLVED]17.2.10 y_p is a particular solution

karush

Well-known member
$y''+2y'=3x$
$y_p = \frac{3}{4}x^2 - \frac{3}{4}x$,
$y(0)=y'(0)=0$

$r^2 + 2r = 0$
$r(r-2)=0$
$r=0$
$r=2$

more steps...

$\textrm{mml final answer}$
$y=\frac{3}{8}\left(1-e^{-2x}\right)+\frac{3}{4}x^2 - \frac{3}{4}x$

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HallsofIvy

Well-known member
MHB Math Helper
Okay, that was the next thing to do! Why? You found and solved the "characteristic equation". Where did you learn that? Didn't you, at the time you learned to find the "characteristic equation", learn why you would want to find it?

I would hope that you learned that, if a and b are distinct roots of the characteristic equation of a linear differential equation with constant coefficients, then $$Ce^{ax}+ De^{bx}$$ is the general solution to the corresponding "homogeneous equation". If you don't know that the whole exercise of finding and solving the characteristic equation is pretty pointless!

And at the same time you learned that (or shortly after) you should have learned that the general solution to a non-homogeneous linear differential equation is the general solution to the associated homogeneous equation plus any particular solution to the entire equation.

karush

Well-known member
I added the y= to the OP but don't know how they got it

I think the Y= is just one step to the original question

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MarkFL

Staff member
I added the y= to the OP but don't know how they got

I think the Y= is just one step to the original question
I was getting ready to reply that you need another initial condition, but you've fixed that. Also, I moved the thread to "Differential Equations" as that's a better fit for this thread.

Once you found the general homogeneous solution $y_h$, then armed with the particular solution $y_p$ you use the principle of superposition to state the general solution is:

$$\displaystyle y(x)=y_h(x)+y_p(x)=c_1+c_2e^{2x}+\frac{3}{4}x^2-\frac{3}{4}x$$

So, what you want to do to find the parameters $c_i$ is to compute $y'(x)$, and the use the two initial conditions to obtain two equations in the two unknowns and solve the system.

Can you proceed?

karush

Well-known member
$\textrm{Differtial equation solution}$
$$y=c_2 \sin(\sqrt{2}x)+c_1\cos(\sqrt{2}x) +\frac{3x}{2}$$
$$y'=\sqrt{2}\sqrt{2}\cos(\sqrt{2}x) -c_1\sqrt{2}\sin(\sqrt{2}x)$$
$\textrm{more steps...}$
$\textrm{mml final answer}$
$y=\frac{3}{8}\left(1-e^{-2x}\right)+\frac{3}{4}x^2 - \frac{3}{4}x$

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karush

Well-known member
$\textrm{set$y=0$and$x=0$}$
$$0=c_2 \sin(\sqrt{2}(0)) +c_1\cos(\sqrt{2}(0)) +\frac{3(0)}{2}$$
$$0=c_2\sqrt{2}\cos(\sqrt{2}(0)) -c_1\sqrt{2}\sin(\sqrt{2}(0))$$
$\textrm{really ???}.$

$\textrm{mml final answer}$
$y=\frac{3}{8}\left(1-e^{-2x}\right)+\frac{3}{4}x^2 - \frac{3}{4}x$

MarkFL

Staff member
I didn't notice before, but your characteristic roots are incorrect...we actually have:

$$\displaystyle r\in\{-2,0\}$$

And so the general solution is:

$$\displaystyle y(x)=c_1+c_2e^{-2x}+\frac{3}{4}x^2-\frac{3}{4}x$$

And thus, we obtain:

$$\displaystyle y'(x)=-2c_2e^{-2x}+\frac{3}{2}x-\frac{3}{4}$$

Using the given initial values, we obtain:

$$\displaystyle y(0)=c_1+c_2=0$$

$$\displaystyle y'(0)=-2c_2-\frac{3}{4}=0$$

From these, we obtain:

$$\displaystyle \left(c_1,c_2\right)=\left(\frac{3}{8},-\frac{3}{8}\right)$$

And so, the solution to the IVP is:

$$\displaystyle y(x)=\frac{3}{8}-\frac{3}{8}e^{2x}+\frac{3}{4}x^2-\frac{3}{4}x=\frac{3}{8}\left(-e^{2x}+2x^2-2x+1\right)$$