Yes, ode45 can be used for higher-order differential equations. You need to convert the higher order equation to a system of first-order equations, then use ode45 on that system.
For example, if you have
... u'' + a·u' + b·u = f
you can define u1 = u, u2 = u' and now you have the system
... (u2)' + a·u2 + b·u1 = f
... (u1)' = u2
Rearranging, this is
... (u1)' = u2
... (u2)' = f - a·u2 - b·u1
ode45 is used to solve each of these. Now, you have a vector (u1, u2) instead of a scalar variable (u). A web search regarding using ode45 on higher-order differential equations can provide additional illumination, including specific examples.
Answer:
34.44 miles
Step-by-step explanation:
Distance covered bty the two trains :
Distance = speed * time
Time = 30 minutes = 0.5hr
Distance of train 1 = 64 * 0.5 = 32 miles
Distance of train 2 = 52 * 0.5 = 26 miles
Included Angle = 72°
Using cosine rule :
a² = b² + c² - 2bcCosA
Let Distance between the trains after 30 minutes = a
a^2 = 32^2 + 26^2 - 2(32)(26) * Cos(72)
a^2 = 1700 - 514.20427
a^2 = 1185.7957
a = sqrt(1185.7957)
a = 34.435
a = 34.44 miles
Answer: 91.44
Step-by-step explanation:
There are 3 feet in a yard, 12 inches in a foot. 2.54 x 12= 30.48
30.48 x 3= 91.44
I hope this helps!
Answer:
Alex has to cycle 
times faster.
Step-by-step explanation:
Let x meters per minute be Alex's initial rate. If Alex’s bike ride to work usually takes 22 minutes, then the distance from home to work is 22x meters.
One morning he found a new bike route that was 5 minutes faster, then he spends 22-5=17 minutes on the way home cycling at the rate x meters per minute. Then the length of the new route is 17x meters.
If he decides to use this new route again and wants to reach home by an additional 2 minutes earlier, then he will spend 17-2=15 minutes to cover 17x meters. So, the new rate will be
Now,
so, Alex has to cycle times faster.
