Find the solution of the differential equation that satisfies the given initial condition.
du/dt= (2t + sec^2(t))/(2u), u(0) = −4
du/dt = (2t + sec^2(t))/(2u)
2u du = (2t + sec^2(t)) dt
u^2 = t^2 + tan(t) + c
16 = 0+0+c
u^2 = t^2 + tan(t) + 16
again for this one do you mean it would be u= - sqrt(t^2+tan(t)+16) if so that is what i have but was just trying to confirm before entering it into the online homework system so i don't seem like an idiot if the professor checks to see how many attempts I've taken on the homework problems.
If you want an explicit function, then you are correct.
To solve the given differential equation and find the solution that satisfies the initial condition, we can use the method of separation of variables.
Step 1: Rearrange the equation to have all the u terms on one side and all the t terms on the other side:
du/u = (2t + sec^2(t)) / (2) dt
Step 2: Integrate both sides of the equation with respect to their respective variables:
∫ du/u = ∫ (2t + sec^2(t)) / 2 dt
Step 3: Evaluate the integrals:
ln|u| = t^2 + tan(t) + C, where C is the constant of integration.
Step 4: Solve for u by taking the exponential of both sides:
|u| = e^(t^2+tan(t)+C)
Step 5: Consider the absolute value signs by separating the equation into two cases:
Case 1: If u > 0, then u = e^(t^2+tan(t)+C)
Case 2: If u < 0, then u = -e^(t^2+tan(t)+C)
Step 6: Use the initial condition u(0) = -4 to find the specific value of the constant C:
When t = 0, u = -4
Plugging these values into the equation gives:
-4 = -e^(0^2+tan(0)+C)
-4 = -e^(C+0)
-4 = -e^C
To isolate e^C, divide both sides by -1 (since the signs cancel):
4 = e^C
Step 7: Substitute the specific value of C back into the equation for u:
Case 1: If u > 0, then u = e^(t^2+tan(t)+ln(4))
Case 2: If u < 0, then u = -e^(t^2+tan(t)+ln(4))
Therefore, the solution to the given differential equation that satisfies the initial condition u(0) = -4 is:
u = e^(t^2+tan(t)+ln(4)) for u > 0
u = -e^(t^2+tan(t)+ln(4)) for u < 0