Could somebody please check these for me?
Solve the differential equation with the given initial solution.
1. (dy/dx)=(-y/3)
for y(0)=10
my answer was y=10e^(-x/3)
2. dP/dt=P+4
for P(0)=100
my answer was P=104(e^t)-4
3. dz/dt=te^z
for z(0)=0
my answer was z=-ln(-.5(t^2)+1)
THANKS
To check the solutions to the given differential equations, you need to differentiate the expressions you found for the dependent variable with respect to the independent variable. Then substitute the initial conditions into the resulting equations and see if they satisfy the original differential equations.
1. The differential equation is (dy/dx) = -y/3.
Taking the derivative of the solution y = 10e^(-x/3) with respect to x:
(dy/dx) = 10e^(-x/3) * (-1/3)
= -10e^(-x/3) / 3
So, the left-hand side (LHS) matches the differential equation.
Substituting the initial condition y(0) = 10 into the solution:
y(0) = 10e^(0) = 10
The right-hand side (RHS) also satisfies the initial condition.
Therefore, your answer y = 10e^(-x/3) is correct.
2. The differential equation is dP/dt = P + 4.
Taking the derivative of the given solution P = 104e^t - 4 with respect to t:
(dP/dt) = 104e^t
So, the LHS matches the differential equation.
Substituting the initial condition P(0) = 100 into the solution:
P(0) = 104e^0 - 4 = 100
The RHS also satisfies the initial condition.
Therefore, your answer P = 104e^t - 4 is correct.
3. The differential equation is dz/dt = te^z.
Taking the derivative of the solution z = -ln(-0.5t^2 + 1) with respect to t:
(dz/dt) = (1 / (-0.5t^2 + 1)) * (-t)
= t / (0.5t^2 - 1)
Please note that the derivative of the natural logarithm requires the chain rule, so make sure you apply it correctly.
It seems like there might be an error in the differentiation of the solution. Double-check your calculations and try differentiating it again. Then substitute the initial condition z(0) = 0 into the derivative and see if it matches the original differential equation.
Please note that finding a closed-form solution for a differential equation may not always be possible. In some cases, numerical methods are used to approximate the solution.