Differential Equations

The derivative $f'(a)$ equals zero.

The function is not differentiable for any other value except at $x = a$.

The function $f(x)$ is continuous at $x = a$.

The function has a local maximum or minimum at $x = a$.

All non-zero values of k result in intersections between particular solutions and their respective slope fields since $ky - k^2$ simplifies into just $-k^2$ showing consistent negative slopes across all values other than zero which generates no slope (flat line).

Any positive value of k creates intersections with their corresponding slope fields at $(0,k)$ since they produce matching slopes through direct substitution into $\frac{dy}{dx} = ky - k^2$ formula when $x = 0$.

No such k exists as any real value would either yield an undefined or unbounded slope at $(0,k)$ due to division by zero or subtraction from itself respectively.

Only $k = 0$ provides such an intersection as it neutralizes both terms resulting in a flat slope field along $y = k$ lines including $y(0)$.

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Using separation of variables on the differential equation.

Performing implicit differentiation on $f(x)$.

Substituting $f(x)$ into the differential equation and simplifying.

Applying partial fractions decomposition technique on $f(x)$.

It indicates that solutions $y$ to this differential equation are unaffected by variations in $x$ related through function $f$ since multiplying anything by zero results in zero.

Zeroing out one term lets us focus more directly on properties intrinsic solely to just those involving variable $x$'s dependence connection via $g$, disregarding contributions that could potentially come from the otherwise multiplied element's secondary curvature rate change. It serves as a principal determinant characterizing the overall behavior of paths taken by particles—physical systems modeled using such DEs when considered in the context of motion under forces described by $g$.

Substitute $P(t)$ into $\frac{dP}{dt}$, simplify completely, then check consistency with original model.

Use partial fractions to decompose $P(t)$, then link back to rate constant k.

Differentiate $P(t)$, compare initial conditions with those implied by k and L.

Incorporate Taylor series expansion around equilibrium point $P=L$.

$f(x) = x^3 - 4$

$f(x) = x^3 - 1$

$f(x) = \frac{x^3}{3} + 4$

$f(x) = x^3 + 1$

The function $y = e^{e^x}$ is not a valid solution because its derivative does not satisfy the initial condition.

The function $y = e^{e^x}$ is not a valid solution since $\ln(y)$ does not exist for all values of $y$.

The function $y = e^{e^x}$ is not a valid solution because it yields a complex number when plugged into the differential equation.

The function $y = e^{e^x}$ is a valid solution as its derivative matches the given differential equation.

$C=3$

$C=-3$

$C=1$

$C=0$

Separation of variables and solving for $C$.

Applying Euler’s Method starting at an initial condition.

Direct substitution of $g(t)$ and its derivative into the differential equation.

Using integration by parts on $\frac{dg}{dt}$.