Solve the following differential equations. 1. (y^2+1)dx = y sec^2 x dy Hint: cos^2 x = (1 + cos 2x) / 2 2. y(ln x - ln y)dx = (x ln x - x ln y - y)dy 3. (6x+1)y^2 dy/dx + 3x^2 + 2y^3 = 0 4. dx/dy = -(4y^2 + 6xy)/(3y^2 + 2x) 5. t dQ/dt + Q = t^4 ln t

          Solve the following differential equations.

1. (y^2+1)dx = y sec^2 x dy Hint: cos^2 x = (1 + cos 2x) / 2

2. y(ln x - ln y)dx = (x ln x - x ln y - y)dy

3. (6x+1)y^2 dy/dx + 3x^2 + 2y^3 = 0

4. dx/dy = -(4y^2 + 6xy)/(3y^2 + 2x)

5. t dQ/dt + Q = t^4 ln t

Solve the following differential equations.

1. (y^2+1)dx = y sec^2 x dy Hint: cos^2 x = (1 + cos 2x) / 2

2. y(ln x - ln y)dx = (x ln x - x ln y - y)dy

3. (6x+1)y^2 dy/dx + 3x^2 + 2y^3 = 0

4. dx/dy = -(4y^2 + 6xy)/(3y^2 + 2x)

5. t dQ/dt + Q = t^4 ln t

Added by Ver-Nica S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/16/2023

Step 1

Given: $(1+\cos{2x})(y+1)dx = y\sec{x} dy$ Divide both sides by $(1+\cos{2x})(y+1)$: $\frac{dx}{dy} = \frac{y\sec{x}}{(1+\cos{2x})(y+1)}$ Now, integrate both sides with respect to $y$: $\int \frac{dx}{dy} dy = \int \frac{y\sec{x}}{(1+\cos{2x})(y+1)} dy$ $x = Show more…

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Solve the following differential equations: 1. (y^2+1)dx = y sec^2 x dy Hint: cos^2 x = (1+cos 2x)/2 2. y(ln x - ln y)dx = (x ln x - x ln y - y)dy 3. (6x+1)y^2 dy/dx + 3x^2 + 2y^3 = 0 4. dx/dy = -(4y^2 + 6xy)/(3y^2 + 2x) 5. t dQ/dt + Q = t^4 ln t