Solve the initial value problem.\\ 8) $\frac{ds}{dt} = 20t(5t^2 - 1)^3$, $s(1) = -1$
Added by Brenda B.
Close
Step 1
First, let's rewrite the given equation in a more standard form: dy/dt = 20t(5t^2 - 13) + s(1) = -1 Now, let's separate the variables by moving all terms involving y to one side and all terms involving t to the other side: dy = (-1 - 20t(5t^2 - 13)) dt Show more…
Show all steps
Your feedback will help us improve your experience
Suman K and 95 other Calculus 1 / AB 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. ds/dt = 20t(5t^2 - 3)^3, s(1) = 2
Adi S.
Solve the initial value problem. $frac{ds}{dt} = 20t(5t^2 + 3)^8$, $s(1) = 2$
Suman 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