By inspection find a particular solution of $y''+2y=−4x$

Question

By inspection find a particular solution of $y''+2y=−4x$.

Write the answer in the form $a_nx^n+\ldots+a_0$ with the highest degree order.

I know how to do this when there's one variable, by finding the value of $y$, but when there are two I'm not sure if I have to find the value of both $x$ and $y$ or ...?

Any help would be appreciated. Thank you!

Answer 1

Here you can use the method of undetermined coefficients, where the solution will be the sum of complementary solution (homogenous solution) and the particular solution.

Let $y=e^{\lambda x}$

$$\frac{d^2}{d^2x}(e^{\lambda x}) + 2e^{\lambda x}= 0$$

I'll leave the solution as an exercise, but you should get some function of $Y$ that requires Euler's identity.

Next, you find the particular solution, often denoted as $y_p$. In this case, the particular solution will take the form $y_p (x) = a_1 + a_2 x$. You can plug this into your differential equation and solve for $a_1$ and $a_2$.

Then let $y = y_h + y_p$ and that is your implicit solution. Solve for any constants with initial conditions as you normally would.

Answer 2

If you attempt to solve the differential equation you want to solve for

$$ y(x) = y_{h} + y_{p} \tag{1} $$

if we solve the homogeneous solve we can look at the characteristic equation

$$ e^{\lambda x}(r^{2} + 2) = 0 \\ r^{2} + 2 = 0 \\ r = \pm \sqrt{2}\tag{2} $$ which gives us

$$ y_{h} = c_{1}\cos(\sqrt{2}x) + c_{2}\sin(\sqrt{2}x) \tag{3} $$

then to find the particular solution we have

$$ y_{p} = c_{1} + c_{2}x \tag{4} $$

to find this we plug it back into the equation

$$ y_{p}^{''}(x) + 2y_{p} = -4x \tag{5} $$

$$ y_{p}^{''}(x) = 0 \\ 2y_{p} = -4x $$ simply solve this

$$ 2c_{1} +2c_{2}x = -4x \implies c_{1} = 0 \tag{6}$$

$$ 2c_{2} = -4 \implies c_{2} = -2 \\ y_{p} = -2x \tag{7} $$

then we get

$$ y = y_{h} +y_{p} = c_{1}\cos(\sqrt{2}x) + c_{2}\sin(\sqrt{2}x) -2x \tag{8}$$

Answer 3

$$y''+2y=−4x\implies (D^2+2)y=-4x$$

Roots of the trial solution are $~±i~\sqrt2~$ .

Complementary function (C.F.) is $$c_1~\cos(\sqrt2~x)+c_2~\sin(\sqrt2~x)\qquad \text{where $~c_1~$ and $~c_2~$are constants.}$$

Particular integral (P.I.) is

P.I. $~=~\frac{1}{D^2+2}~(-4x)$

$~=~-4~\frac{1}{D^2+2}~x$

$~=~-4~\cdot\frac{1}{2}~(1+\frac{D^2}{2})^{-1}~x$

$~=~-2~(1-\frac{D^2}{2}+\cdots)~x$

$~=~-2~x$

Hence the general solution is$$y(x)=\text{C.F.}~+~\text{P.I.}$$ $$y(x)=c_1~\cos(\sqrt2~x)+c_2~\sin(\sqrt2~x)~-2~x$$where $~c_1~$ and $~c_2~$are constants.

Answer 4

Try to absorb the right side into the lowest derivative term on the left. That is, take $u(x)=y(x)+2x$ as new variable. Then $$ u''(x)+2u(x)=y''(x)+2y(x)+4x=0 $$ and from the solution $u=0$ we get the particular solution $y(x)=-2x$.