Question

Find the solution to the differential equation

dz/dt = 7te^2 z that passes through the origin.
z=?

Answer 1

Since
`\(z = 0\)` is a solution, the solution to the given differential equation that passes through the origin is:
`\[ z = 0 \]`

Step-by-step explanation:

To solve the differential equation
`\((dz)/(dt) = 7te^(2t)z\)`that passes through the origin, we can separate variables and integrate.

Separate variables:

`\[(1)/(z) \, dz = 7te^(2t) \, dt\]`

Now integrate both sides:

`\[ \int (1)/(z) \, dz = \int 7te^(2t) \, dt \]`

Integrating the left side gives the natural logarithm:

`\[ \ln|z| = \int 7te^(2t) \, dt \]`

Now integrate the right side. We can use integration by parts, where
`\(u = t\)` and
`\(dv = 7e^(2t) \, dt\):`

`\[ \ln|z| = \int 7te^(2t) \, dt = (7)/(2)te^(2t) - (7)/(2)\int e^(2t) \, dt \]`

`\[ \ln|z| = (7)/(2)te^(2t) - (7)/(4)e^(2t) + C \]`

Here,
`\(C\)` is the constant of integration.

Now, we need to exponentiate both sides to solve for
`\(z\)`:

`\[ |z| = e^{(7)/(2)te^(2t) - (7)/(4)e^(2t) + C} \]`

Since we are looking for a solution that passes through the origin, we know that
`\(z(0) = 0\).` Substituting
`\(t = 0\)`, we get:

`\[ |z(0)| = e^(C) \]`

Since
`\(z(0) = 0\)`, we have
`\(|z(0)| = 0\)`, so
`\(e^C = 0\)`. But the exponential function is never zero, so
`\(C\)` must be
`\(-\infty\)`.

Now, substitute
`\(C = -\infty\)` back into the expression for
`\(z\)` :

`\[ z = \pm e^{(7)/(2)te^(2t) - (7)/(4)e^(2t) - \infty} \]`