# What is a solution to the differential equation dy/dt=e^t(y-1)^2?

Apr 4, 2018

The General Solution is:

$y = 1 - \frac{1}{{e}^{t} + C}$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{t} {\left(y - 1\right)}^{2}$

We can collect terms for similar variables:

$\frac{1}{y - 1} ^ 2 \setminus \frac{\mathrm{dy}}{\mathrm{dt}} = {e}^{t}$

Which is a separable First Order Ordinary non-linear Differential Equation, so we can "separate the variables" to get:

$\int \setminus \frac{1}{y - 1} ^ 2 \setminus \mathrm{dy} = \int {e}^{t} \setminus \mathrm{dt}$

Both integrals are those of standard functions, so we can use that knowledge to directly integrate:

$- \frac{1}{y - 1} = {e}^{t} + C$

And we can readily rearrange for $y$:

$- \left(y - 1\right) = \frac{1}{{e}^{t} + C}$

$\therefore 1 - y = \frac{1}{{e}^{t} + C}$

$y = 1 - \frac{1}{{e}^{t} + C}$

Apr 4, 2018

$y = - \frac{1}{{e}^{t} + C} + 1$

#### Explanation:

This is a separable differential equation, which means it can be written in the form:

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot f \left(y\right) = g \left(x\right)$

It can be solved by integrating both sides:

$\int \setminus f \left(y\right) \setminus \mathrm{dy} = \int \setminus g \left(x\right) \setminus \mathrm{dx}$

In our case, we first need to separate the integral into the right form. We can do this by dividing both sides by ${\left(y - 1\right)}^{2}$:

$\frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{1}{y - 1} ^ 2 = {e}^{t} \cancel{{\left(y - 1\right)}^{2} / {\left(y - 1\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{1}{y - 1} ^ 2 = {e}^{t}$

Now we can integrate both sides:

$\int \setminus \frac{1}{y - 1} ^ 2 \setminus \mathrm{dy} = \int \setminus {e}^{t} \setminus \mathrm{dt}$

$\int \setminus \frac{1}{y - 1} ^ 2 \setminus \mathrm{dy} = {e}^{t} + {C}_{1}$

We can solve the left hand integral with a substitution of $u = y - 1$:

$\int \setminus \frac{1}{u} ^ 2 \setminus \mathrm{du} = {e}^{t} + {C}_{1}$

$\int \setminus {u}^{-} 2 \setminus \mathrm{du} = {e}^{t} + {C}_{1}$

${u}^{-} \frac{1}{- 1} + {C}_{2} = {e}^{t} + {C}_{1}$

Resubstituting (and combining constants) gives:

$- \frac{1}{y - 1} = {e}^{t} + {C}_{3}$

Multiply both sides by $y - 1$:

$- 1 = \left({e}^{t} + {C}_{3}\right) \left(y - 1\right)$

Divide both sides by ${e}^{t} + {C}_{3}$:

$- \frac{1}{{e}^{t} + {C}_{3}} = y - 1$

$y = - \frac{1}{{e}^{t} + C} + 1$