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hi i have a problem for this Differential Equations : $$ \frac{d^{3}y}{dx^3} - 9\frac{dy}{dx} = 10 - 4x $$ i know first we must solve the homogeneous equation: and my result is : $C_1 + C_2e^{3x} + C_3e^{-3x}$

i'm really confused to find yp from $10 - 4x$, my friend said it must : $(ax+b)x$ my question is :
1. why it's not $c - (ax + b)$,
2. if it same with $c1$ from yh(homogeneous solution ) why we not multiply $c$ with x? so $yp = cx - (ax + b) $
3. and why $(ax+b)$ must be multiplied with $x$??

can someone explain me? thanks for anyone helped me.

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  • $\begingroup$ If you want to know why you can't use $y_p=c-(ax+b)$ or $y_p=cx-(ax+b)$ then you should try to use one of those, and see what happens. When you see why it doesn't work, you will understand why it is necessary to try $y_p=(ax+b)x$. $\endgroup$ – Gerry Myerson Jan 28 '13 at 11:43
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It is straightforward to see that if a polynomial solves the inhomogeneous ODE $$\tag{ODE} y'''-9y'=10-4x, $$ it must be of degree $2$. Hence, we have $$ p'''(x)-9p'(x)=10-4x \iff -9p'(x)=10-4x \iff p'(x)=-\frac{10}{9}+\frac49x, $$ i.e. $$ p(x)=-\frac{10}{9}x+\frac29x^2 $$ Setting $$\tag{2} z=y-p,\ u=z' $$ we see that $z$ solves the homogeneous equation $$\tag{ODE'} z'''-9z'=0, $$ and therefore $u$ solves the 2nd order ODE $$ u''-9u=0 $$ whose solution is given by $$ u(x)=-3c_1e^{-3x}+3c_2e^{3x}, \ c_1,c_2 \in \mathbb{R}. $$ It follows that $$ z(x)=c_3+c_1e^{-3x}+c_2e^{3x},\ c_3 \in \mathbb{R} $$ and $$ y(x)=c_3-\frac{10}{9}x+\frac29x^2+c_1e^{-3x}+c_2e^{3x}. $$

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Related problems: (I). We use the annihilator method. Let $D=\frac{d}{dx}$. Now, apply $D$ twice to both sides of the differential equation to transform it to a homogeneous differential equation, so we have

$$ D^2(D^3-9D)y=D^2(10-4x)=0\implies D^3(D^2-9)y=0. $$

The last equation gives you the auxiliary equation of the new homogeneous differential equation which allow us to write the general solution of the new homogeneous differential equation

$$ m^3(m^2-9)=0 \implies m=0,0,0,3,-3,$$

which gives the general solution

$$ y(x) = c_1+c_2x+c_3x^3+c_4e^{3x}+c_5e^{-3x}=(c_1+c_4e^{3x}+c_5e^{-3x})+c_2x+c_3x^2 \rightarrow(1).$$

You can recognize that the expression between the brackets in equation $(1)$ corresponds to the solution of the original homogeneous differential equation

$$ \frac{d^{3}y}{dx^3} - 9\frac{dy}{dx} = 0. $$

So, the other terms in $(1)$ tell you about the form of $y_p$ of the original differential equation which you assume as

$$ y_p=Ax +Bx^2. $$

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Inhomogeneous solution here will be of the form

$$y_p(x) = a (10-4 x)^2 + b$$

such that $y_p(0) = 0$ (subsequent values of $y_p'(0)$ and $y_p''(0)$ will have to be taken into account into overall initial conditions on solution $y$). The reason it takes a quadratic form is that you have a $y'$ but no $y$ in your equation.

Putting this form into the equation and solving for $a$ and $b$ (from the initial condition), I get

$$y_p(x) = \frac{1}{72} (10-4 x)^2 - \frac{25}{18}$$

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