Question

For the differential equation y′′+4y′+4y=x3 Part 1: Solve the homogeneous equation The differential operator for the homogeneous equation is List the complementary functions (the functions that make up the complementary solution) . When you get this answer correct it will give you the format for the complementary solution that you must use below. Part 2: Find the particular solution To solve the non-homogeneous differential equation, we look for functions annihilated by the differential operator (a multiple of the operator given above) Therefore the particular solution must be made up of the functions Substituting these into the differential equation, we find the particular solution is Part 3: Solve the non-homogeneous equation y′′+4y′+4y=x3 has general solution (remember to use the format I gave you in your correct answer to the complementary functions above) Now that we have the general solution solve the IVP y(0)=−6 y′(0)=8

Answer 1

1) The homogeneous equation is

$y''+4y'+4y=0$

with characteristic equation

$r^2+4r+4=(r+2)^2=0\implies r=-2$

which means we have the characteristic solution

$y_c=C_1e^{-2x}+C_2xe^{-2x}$

The functions that make up this solution are $e^{-2x}$ and $xe^{-2x}$.

2) Several ways to do this. Using the method of undetermined coefficients, we suppose the particular takes the form

$y_p=a_3x^3+a_2x^2+a_1x+a_0$

so that its derivatives are

${y_p}'=3a_3x^2+2a_2x+a_1$

${y_p}''=6a_3x+2a_2$

Then substituting these into the ODE gives

$(6a_3x+2a_2)+4(3a_3x^2+2a_2x+a_1)+4(a_3x^3+a_2x^2+a_1x+a_0)=x^3$

$4a_3x^3+(4a_2+12a_3)x^2+(4a_1+8a_2+6a_3)x+(4a_0+4a_1+2a_2)=x^3$

Matching up the coefficients gives us the system

$\begin{cases}4a_3=1\\4a_2+12a_3=0\\4a_1+8a_2+6a_3=0\\4a_0+4a_1+2a_2=0\end{cases}\implies a_3=\dfrac14,a_2=-\dfrac34,a_1=\dfrac98,a_0=-\dfrac34$

so the particular solution would be

$y_p=\dfrac{x^3}4-\dfrac{3x^2}4+\dfrac{9x}8-\dfrac34$

3) The general solution is simply

$y=y_c+y_p$

Given the initial values $y(0)=-6$ and $y'(0)=8$, we get

$-6=C_1-\dfrac34\implies C_1=-\dfrac{21}4$

$8=-2C_1+C_2+\dfrac98\implies C_2=-\dfrac{29}8$

giving a particular solution to the IVP of

$y=-\dfrac{21}4e^{-2x}-\dfrac{29}8xe^{-2x}+\dfrac{x^3}4-\dfrac{3x^2}4+\dfrac{9x}8-\dfrac34$

Answer 2

By  finding complementary solution and particular solution  we got the

the solution of IVP as $y=-\frac{21}{4} e^{-2 x}-\frac{29}{8} x e^{-2 x}+\frac{x^{3}}{4}-\frac{3 x^{2}}{4}+\frac{9 x}{8}-\frac{3}{4}$

What is a differential equation?

An equation of function and their derivatives is called differential equation .

Given differential equation

$y^{\prime \prime}+4 y^{\prime}+4 y=x^3$

Homogenous equation can be written as

$y^{\prime \prime}+4 y^{\prime}+4 y=0$

characteristic equation of  this differential equation can be written as

$r^{2}+4 r+4=0\\\Rightarrow (r+2)^{2}=0\\\Rightarrow r=-2,-2$

Roots are repeating

Hence we can write complementary solution as

$y_{C}=C_{1} e^{-2 x}+C_{2} x e^{-2 x}$

The functions that are making up this solution are  $e^{-2 x}$ and $x e^{-2 x} .$

Now particular solution

Let suppose that particular solution is of the form

$y_{p}=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$

So

$&y_{p}^{\prime}=3 a_{3} x^{2}+2 a_{2} x+a_{1} \\\\&y_{p}^{\prime \prime}=6 a_{3} x+2 a_{2}$

Putting these values in given differential equation

$&4 a_{3} x^{3}+\left(4 a_{2}+12 a_{3}\right) x^{2}+\left(4 a_{1}+8 a_{2}+6 a_{3}\right) x+\left(4 a_{0}+4 a_{1}+2 a_{2}\right)=x^{3}$

Now by comparing the coefficient of both sides

$4 a_{3}=1 \ \ \Rightarrow a_{3}=\frac{1}{4}\\\\4 a_{2}+12 a_{3}=0 \ \ \Rightarrow 4 a_{2}+3=0 \ \ \ \ \Rightarrow a_{2}=\frac{-3}{4} \ \ \ \\\\\4 a_{1}+8 a_{2}+6 a_{3}=0 \ \ \ \Rightarrow 4 a_{1}-6+6 \times \frac{1}{4}=0 \\\\ \Rightarrow a_{1}=\frac{9}{8}\\\\4 a_{0}+4 a_{1}+2 a_{2}=0 \ \ \ \Rightarrow a_{0}=-3/4$

Hence we can write  particular solution as

$y_{p}=\frac{x^{3}}{4}-\frac{3 x^{2}}{4}+\frac{9 x}{8}-\frac{3}{4}$

We know general solution

$y= y_{c}+ y_{p}$

$y=e^{-2 x}+C_{2} x e^{-2 x}+\frac{x^{3}}{4}-\frac{3 x^{2}}{4}+\frac{9 x}{8}-\frac{3}{4}$

Now given initial values $y(0)=-6$ and $y^{\prime}(0)=8$

So

$y=-\frac{21}{4} e^{-2 x}-\frac{29}{8} x e^{-2 x}+\frac{x^{3}}{4}-\frac{3 x^{2}}{4}+\frac{9 x}{8}-\frac{3}{4}$

By  finding complementary solution and particular solution  we got the

the solution of IVP as $y=-\frac{21}{4} e^{-2 x}-\frac{29}{8} x e^{-2 x}+\frac{x^{3}}{4}-\frac{3 x^{2}}{4}+\frac{9 x}{8}-\frac{3}{4}$