Given a separable differential equation, find an implicit formula for the general solution to that equation. If possible, arrange that formula into an explicit solution. Given initial conditions on a solution, find the (implicit or explicit) formula for the particular solution corresponding to those conditions.
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Consider the differential equation \(\dot{y} = 6y^2t\,.\)
- Solve this differential equation.
- Solve this differential equation subject to the condition that \(y = \frac{1}{9}\) when \(t=1\,.\)
- Solve the initial value problem: What is a formula \(y = f(t)\) for \[\dot{y} = \frac{\mathrm{e}^t}{y} \quad\text{subject to}\quad f(0)=1\,?\]
- Given that \(y=2\mathrm{e}\) when \(t = 3,\) find an explicit particular solution to the differential equation \[ \frac{y'}{x} = \mathrm{e}^{x-\ln\left(y^2\right)} \,. \]
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Find a general solution to each of the following differential equations.
If possible, arrange the solution into an explicit formula \(y = f(t).\)
\[ \dot y = t\mathrm{e}^{t^2-\ln\left(y^2\right)}\]\[ t^2\dot y = \sec(y) - \dot y \]\[ \sin^2(y)\dot y = (1-\dot y)\cos^2(y) \]\[ t\dot y = \sqrt{t^2 - (ty)^2} \]\[ y\dot{y} + y^2 = 3ty\dot{y}+1 \]\[ \dot{y}\cot^2(t) + \tan(y) = 0 \]\[ t\dot{y} - 27\dot{y} = 3(y-9) \]\[ (ty)^2+2y^2 + \dot{y} = 0 \]\[ 2\cos(t) = 3t^2-\dot{y} \]\[ \dot{y}-6y = 4 \]\[ t\dot{y} - 2\dot{y} = 2(y-4) \]
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Find the particular solution to each of the following differential equations
corresponding to the given initial conditions.
If possible, arrange the solution into an explicit formula \(y = f(t).\)
\[ \frac{1}{t}\sec^2(y)\dot{y} = 1 \quad\text{subject to}\quad y=1 \text{ when } t=0 \]\[ \mathrm{e}^t - y\dot{y} = 0 \quad\text{subject to}\quad y=3 \text{ when } t=0 \]\[ \ddot{y} = t^2 + 3 \quad\text{where}\quad y=-4 \text{ and } \dot{y}=2 \text{ when } t=0 \]