Review 6: Now You Try II
Exercise 1: Consider the differential equation
dydx=7y2x3.
Show that 
y=−174x4+C is a solution to the differential equation, where 
C is a constant.
Solution:
▶ Show
We need to plug in
 "y(x)")
y(x) into the left hand side and the right hand side of the equation and check to see that they are equal.
On the left hand side, we get
%255E2%257D%2520.%250A%255Cend%257Bequation%257D "\begin{equation}\label{eq:left3}
\frac{dy}{dx} = \frac{7x^3}{\left(\frac{7}{4}x^4+C\right)^2} .
\end{equation}")
dydx=7x3(74x4+C)2.
On the right hand side, we get
%255E2%257D%255Ccdot%2520x%255E3.%2520%250A%255Cend%257Bequation%257D "\begin{equation}\label{eq:right3}
7y^2x^3 =7\cdot \frac{1}{\left(\frac{7}{4}x^4+C\right)^2}\cdot x^3.
\end{equation}")
7y2x3=7⋅1(74x4+C)2⋅x3.Since the left hand side and the right hand side are equal, we conclude that y(x) is a solution to the differential equation.
Exercise 2: Consider the differential equation
%2520%252B%2520x(t)%2520%253D%25200.%250A%2520%255Cend%257Bequation%257D "\begin{equation}\label{eq:ed4}
x''(t) + x(t) = 0.
\end{equation}")
x″(t)+x(t)=0.
Show that %2520%253D%2520A%2520%255Ccos(t)%252B%2520B%255Csin(t) "x(t) = A \cos(t)+ B\sin(t)")
x(t)=Acos(t)+Bsin(t) is a solution to the above differential equation, where 
A and 
B are constants.
Solution:
▶ Show
We need to plug in
 "x(t)")
x(t) into the left hand side and the right hand side of the equation and check to see that they are equal. Since
%2520%253D%2520-A%255Ccos(t)%2520-%2520B%255Csin(t)%252C "x''(t) = -A\cos(t) - B\sin(t),")
x″(t)=−Acos(t)−Bsin(t), we get on the left hand side
%252B%2520x(t)%2520%253D%2520-A%255Ccos(t)%2520-%2520B%255Csin(t)%2520%252B%2520A%255Ccos(t)%252BB%255Csin(t)%253D0.%250A%255Cend%257Bequation%257D "\begin{equation}\label{eq:left4}
x''(t)+ x(t) = -A\cos(t) - B\sin(t) + A\cos(t)+B\sin(t)=0.
\end{equation}")
x″(t)+x(t)=−Acos(t)−Bsin(t)+Acos(t)+Bsin(t)=0.On the right hand side, we simply have 0.
Since the left hand side and the right hand side are equal, we conclude that
x(t) is a solution to the differential equation.
Exercise 3: Consider the differential equation we had on the first page (below)
.%250A%2520%255Cend%257Bequation%257D "\begin{equation}\label{eq:ed5}
\frac{dA}{dt} = kA(t).
\end{equation}")
dAdt=kA(t).
Show that %2520%253D%2520Ce%255E%257Bkt%257D "A(t) = Ce^{kt}")
A(t)=Cekt is a solution to the differential equation above, where C is a constant.
Solution:
▶ Show
We need to plug in
 "A(t)")
A(t) into the left hand side and the right hand side of the equation and check to see that they are equal.
On the left hand side, we get
\begin{equation}\label{eq:left3}
\frac{dA}{dt} = Cke^{kt} .
\end{equation}
On the right hand side, we get
\begin{equation}\label{eq:right3}
kA =kCe^{kt}.
\end{equation}
Since the left hand side and the right hand side are equal, we conclude that A(t) is a solution to the differential equation.