\frac{dv}{dt} - 2tv = 3t^2 e^{t^2}, v(0) = 8.
Added by M-Nica C.
Close
Step 1
The product rule states that if we have two functions u(t) and v(t), then the derivative of their product is given by: (d/dt)(u(t)v(t)) = u'(t)v(t) + u(t)v'(t) In this case, u(t) = 2t and v(t) = e^(-t). Taking the derivative of u(t) and v(t), we have: u'(t) = Show more…
Show all steps
Your feedback will help us improve your experience
Brent Burkett and 75 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Solve the initial-value problem: dv/dt - 2tv = 6t^5e^t^2, v(0) = 7
Adi S.
Solve the initial value problem du/dt = e^(2u+3t) u(0) = 9
Sri K.
Solve the initial value problems. $$\frac{d v}{d t}=\frac{8}{1+t^{2}}+\sec ^{2} t, \quad v(0)=1$$
Applications of Derivatives
Antiderivatives
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Watch the video solution with this free unlock.
EMAIL
PASSWORD