Question

2. Are the following differential equations exact? If not, try t, y, and y^2 as possible integrating factors. (a) 2(t^3 + 1) dy + 3yt^2 dt = 0 (b) 4y^3t dy + (2y^4 + 3t) dt = 0 3. By applying the four-step procedure to the general exact differential equation M dy + N dt = 0, derive the following formula for the general solution of an exact differential equation: ? M dy + ? N dt - ? (?/?t ? M dy) dt = c

          2. Are the following differential equations exact? If not, try t, y, and y^2 as possible integrating factors.
(a) 2(t^3 + 1) dy + 3yt^2 dt = 0
(b) 4y^3t dy + (2y^4 + 3t) dt = 0

3. By applying the four-step procedure to the general exact differential equation M dy + N dt = 0, derive the following formula for the general solution of an exact differential equation:
? M dy + ? N dt - ? (?/?t ? M dy) dt = c

2. Are the following differential equations exact? If not, try t, y, and y^2 as possible integrating factors.
(a) 2(t^3 + 1) dy + 3yt^2 dt = 0
(b) 4y^3t dy + (2y^4 + 3t) dt = 0

3. By applying the four-step procedure to the general exact differential equation M dy + N dt = 0, derive the following formula for the general solution of an exact differential equation:
? M dy + ? N dt - ? (?/?t ? M dy) dt = c

Added by Mark T.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/08/2023

Step 1

The given equation is: $$2(t^8 + 1) \frac{dy}{dt} + 3yt^2 \frac{dt}{dt} = 0$$ Let $M = 2(t^8 + 1)$ and $N = 3yt^2$. Then, we need to check if $\frac{\partial M}{\partial t} = \frac{\partial N}{\partial y}$. $$\frac{\partial M}{\partial t} = Show more…

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2. Are the following differential equations exact? If not, try t, y, and y^2 as possible integrating factors. (a) 2(t^3 + 1) dy + 3yt^2 dt = 0 (b) 4y^3t dy + (2y^4 + 3t) dt = 0 3. By applying the four-step procedure to the general exact differential equation M dy + N dt = 0, derive the following formula for the general solution of an exact differential equation: ∫ M dy + ∫ N dt - ∫ (∂/∂t ∫ M dy) dt = c