Assignment-daixieTM为您提供佛罗里达大学University of Florida MAP 4305 Differential Equations微分方程代写代考和辅导服务！
Instructions:
Differential equations are mathematical equations that describe how one or more variables change with respect to one another. Specifically, they involve derivatives or differentials, which represent rates of change.
Differential equations can be used to model a wide range of phenomena in physics, engineering, biology, economics, and many other fields. For example, they can be used to describe the motion of objects under the influence of forces, the spread of disease in a population, the growth of a population of organisms, the behavior of electrical circuits, and the flow of fluids.
Solving differential equations can be challenging, as there may not be an exact formula that provides a solution. Instead, various methods such as separation of variables, substitution, and numerical methods can be used to approximate solutions. Differential equations are an important topic in mathematics and have applications in many scientific and engineering fields.
Find a solution of
$$
y^{\prime}=x^3
$$
such that $y(1)=2$.
Solution At the beginning of this section we saw that the solutions of $y^{\prime}=x^3$ are
$$
y=\frac{x^4}{4}+c
$$
To determine a value of $c$ such that $y(1)=2$, we set $x=1$ and $y=2$ here to obtain
$$
2=y(1)=\frac{1}{4}+c, \text { so } \quad c=\frac{7}{4} .
$$
Therefore the required solution is
$$
y=\frac{x^4+7}{4}
$$
(a) Find the general solution of
$$
y^{\prime \prime}-2 y^{\prime}+y=-3-x+x^2
$$
(b) Solve the initial value problem
$$
y^{\prime \prime}-2 y^{\prime}+y=-3-x+x^2, \quad y(0)=-2, \quad y^{\prime}(0)=1
$$
SOLUTION(a) The characteristic polynomial of the complementary equation
$$
y^{\prime \prime}-2 y^{\prime}+y=0
$$
is $r^2-2 r+1=(r-1)^2$, so $y_1=e^x$ and $y_2=x e^x$ form a fundamental set of solutions of the complementary equation. To guess a form for a particular solution, we note that substituting a second degree polynomial $y_p=A+B x+C x^2$ into the left side will produce another second degree polynomial with coefficients that depend upon $A, B$, and $C$. The trick is to choose $A, B$, and $C$ so the polynomials on the two sides of (5.3.10) have the same coefficients; thus, if
$$
y_p=A+B x+C x^2 \text { then } y_p^{\prime}=B+2 C x \text { and } y_p^{\prime \prime}=2 C \text {, }
$$
so
$$
\begin{aligned}
y_p^{\prime \prime}-2 y_p^{\prime}+y_p & =2 C-2(B+2 C x)+\left(A+B x+C x^2\right) \
& =(2 C-2 B+A)+(-4 C+B) x+C x^2=-3-x+x^2 .
\end{aligned}
$$
Equating coefficients of like powers of $x$ on the two sides of the last equality yields
$$
\begin{aligned}
C & =1 \
B-4 C & =-1 \
A-2 B+2 C & =-3,
\end{aligned}
Find a particular solution of
$$
y^{\prime \prime}-7 y^{\prime}+12 y=4 e^{2 x} .
$$
Then find the general solution.
Solution Substituting $y_p=A e^{2 x}$ for $y$ in will produce a constant multiple of $A e^{2 x}$ on the left side , so it may be possible to choose $A$ so that $y_p$ is a solution. Let’s try it; if $y_p=A e^{2 x}$ then
$$
y_p^{\prime \prime}-7 y_p^{\prime}+12 y_p=4 A e^{2 x}-14 A e^{2 x}+12 A e^{2 x}=2 A e^{2 x}=4 e^{2 x}
$$
if $A=2$. Therefore $y_p=2 e^{2 x}$ is a particular solution. To find the general solution, we note that the characteristic polynomial of the complementary equation
$$
y^{\prime \prime}-7 y^{\prime}+12 y=0
$$
is $p(r)=r^2-7 r+12=(r-3)(r-4)$, so $\left{e^{3 x}, e^{4 x}\right}$ is a fundamental set of solutions of (5.4.3). Therefore the general solution is
$$
y=2 e^{2 x}+c_1 e^{3 x}+c_2 e^{4 x}
$$
