Verify that \( y(t) = -\cos t + 6 \) is the solution to
\[ \text{IVP} \quad \underbrace{y' = \sin t}_{\text{ODE}}, \quad \underbrace{y(0) = 5}_{\text{initial condition}} \]
\[ \frac{dy}{dt} = \frac{d}{dt}(-\cos t + 6) = \sin t \]
\[ y(0) = -\cos(0) + 6 = 5 \]
Rewrite the differential equation:
\[ \frac{dy}{dt} = \sin t \]
Integrate both sides (or find the antiderivative) to obtain the general solution:
\[ y(t) = \int \sin t \, dt = -\cos t + C \]
The equation \( y(t) = -\cos t + C \) represents the general solution (a 1-parameter family of solutions).
Substitute the initial condition \( y(0) = 5 \) to solve for the constant \( C \) and find the particular solution:
\[ 5 = y(0) = -\cos(0) + C \implies C = 6 \]
Thus, the unique solution to the IVP is:
\[ y(t) = -\cos t + 6 \]
To solve the equation, integrate both sides with respect to \( t \) (or find the antiderivative of \( f(t) \)):
\[ y(t) = \int f(t) \, dt + c \]
\( \leadsto \) General solution (or 1-parameter family of solutions).
Suppose you are given an initial condition \( y(0) = y_0 \):
Find the constant \( c \) by plugging \( y(0) = y_0 \) into the general solution.
Integrate
\[ y(x) = \int (3x + \cos x) \, dx + C \] \[ y(x) = \frac{3x^2}{2} + \sin x + C \quad \Bigg\} \text{ General Solution} \] \[ 5 = y(0) = 3 \cdot \frac{0^2}{2} + \sin(0) + C \implies C = 5 \] \[ y(x) = \frac{3x^2}{2} + \sin x + 5 \implies \text{Solution to IVP} \]
Integrate twice to find the general solution
\[ y' = \int 3x \, dx + c = \frac{3x^2}{2} + c \] \[ y = \int \left( \frac{3x^2}{2} + c \right) dx + d \] \[ y = \frac{x^3}{2} + cx + d \]
General solution, or 2-parameter family of solutions
Example:
\[ y'' = 3x + \cos x \]
integrate
\[ y' = \int 3x + \cos x \, dx + c = \frac{3x^2}{2} + \sin x + c \]
integrate
\[ y = \int \frac{3x^2}{2} + \sin x + c \, dx + d \] \[ = \frac{x^3}{2} - \cos x + cx + d \]
To find particular solution, we need two conditions
typically this is \( y(0), \, y'(0) \)
\[ y'' = f(t), \quad y(0) = y_0, \quad y'(0) = y_1 \]
General solution: \( y(x) = \frac{x^3}{2} + cx + d \)
\[ y(0) = 4 \leadsto 4 = y(0) = \frac{0^3}{2} + c(0) + d \Rightarrow d = 4 \]
\[ y'(0) = -3 \leadsto y'(x) = 3\frac{x^2}{2} + c \]
\[ -3 = y'(0) = 0 + c \Rightarrow c = -3 \]
Solution is
\[ y = \frac{x^3}{2} - 3x + 4 \]
\( x(t) = \) position
\( v(t) = \) velocity \( \rightsquigarrow v(t) = x'(t) \)
\( a(t) = \) acceleration \( \rightsquigarrow a(t) = v'(t) = x''(t) \)
Newtons Second Law:
\[ F(t) = m \cdot a(t) \] \[ a(t) = \frac{F(t)}{m} \] \[ x''(t) = a(t) = \frac{F(t)}{m} \]
Suppose \(x(0) = x_0\), \(v(0) = v_0\) are initial position and velocity
\[ v'(t) = a(t) = a \]
integrate
\[ v(t) = at + c \quad \left. \begin{array}{l} v(0) = v_0 \end{array} \right\} \Rightarrow c = v_0 \] \[ v(t) = at + v_0 \] \[ x'(t) = v(t) = at + v_0 \]
integrate
\[ x(t) = \int at + v_0 \, dt + d = \frac{1}{2} a t^2 + v_0 t + d \]
plug in \(x(0) = x_0\)
\[ x(t) = \frac{1}{2} a t^2 + v_0 t + x_0 \]
eg: Velocity \(v(t)\) is given according to below graph, plot position \(x(t)\) and acceleration \(a(t)\) (assume \(x(0)=0\))
\[\int 5 \, dt = 5t\]
A ball is dropped from the top of a 400ft building
\[ a(t) = \text{acceleration due to gravity} = -9.8\,\text{m/s}^2 = -32\,\text{ft/s}^2 \]
\[ x(0) = 400\,\text{ft} \]
① What is the time when
\[ x(t) = 0 \]
② Once you know the Answer for ① find \( v(t) \)
initial velocity is \( v(0) = 0 \)
IVP:
\[ x'' = -32, \quad x(0) = 400, \quad x'(0) = v(0) = 0 \]
Integrate:
\[ x'(t) = -32t + c \quad \text{w/} \quad 0 = x'(0) = -32(0) + c \implies c = 0 \] \[ x'(t) = -32t \]
Integrate:
\[ x(t) = -16t^2 + d \quad \text{w/} \quad 400 = -16(0)^2 + d \implies d = 400 \] \[ x(t) = -16t^2 + 400 \]
\[ \implies -16t^2 + 400 = 0 \] \[ t^2 = \frac{400}{16} = 25 \] \[ t = 5\text{ sec.} \]
\[ v(t) = x'(t) = -32t \] \[ v(5) = -160\text{ ft/sec.} \]