find the solution of the following IVPs.
i) x^2cosxdx+x(1-6y^2)dy=0, y(pi)=0
ii) (dQ/dt)+2/(10+2t)*Q=4, Q(2)=100

Question

find the solution of the following IVPs.
i) x^2cosxdx+x(1-6y^2)dy=0, y(pi)=0
ii) (dQ/dt)+2/(10+2t)*Q=4, Q(2)=100

kc_kennylau · Answer

$$\mbox{i) }x^2\cos xdx+x(1-6y^2)dy=0,y(\pi)=0$$$$\mbox{ii) }\frac{dQ}{dt}+\frac2{(10+2t)Q}=4,Q(2)=100$$

Code:
x^2\cos xdx+x(1-6y^2)dy=0,y(\pi)=0
\frac{dQ}{dt}+\frac2{(10+2t)Q}=4,Q(2)=100

kc_kennylau · Answer

@Chineseboy15 maybe you can help him? :)

anonymous · Answer

Sorry, I cannot. I have not learnt it very well.

kc_kennylau · Answer

@bambede76 sorry I cannot help you too :(

anonymous · Answer

thank you Chineseboy15 and kc_kennylau for your concern. But sad you cannot help me out.

anonymous · Answer

According to Newton’s Law of Cooling, an object will cool (or warm) at a rate proportional to the difference in temperature between the object and its environment. If the temperature of the object is known at some time, say T = T0 at time t = 0, with an environmental temperature of Te, then a formula for the temperature T(t) can be solved by solving the initial-value problem

					  ( dT )/( dt )=k( T- T_e ), 	 T(0)= T_0  

At 10:30 AM, detectives discover a dead body in a room and measure the body’s temperature at 26°C. One hour later, the body’s temperature had dropped to 24.8°C. Determine the time of death (when the body temperature was a normal 37°C), assuming that the temperature in the room was held to constant at 20°C.