Differential Equations - Separable Variables

Solve the general solution for the ODE:

$\frac{dy}{dx}=x^8y$

$\ln(y) = \frac{x^9}{9}+C$

where C is an arbitrary constant

$y = \frac{x^9}{9}+C$

where C is an arbitrary constant

$\ln(y) = x^8+C$

where C is an arbitrary constant

$\ln(y) = e^\frac{x^9}{9}+C$

where C is an arbitrary constant

Explanation

First the differential equation can be separated to:

$\frac{dy}{y} = x^8dx$

And then integrated simply to:

$\ln(y) = \frac{x^9}{9}+C$

Solve the following differential equation

$\frac{dy}{dx}=3x^2y$

$y=Ce^x^{^{3}}$

$y=e^x^{^{3}}$

$y=e^x^{^{3}+C}$

$ln(y)=e^x^{^{3}}$

Explanation

So this is a separable differential equation. The first step is to move all of the x terms (including dx) to one side, and all of the y terms (including dy) to the other side.

So the differential equation we are given is: $\frac{dy}{dx}=3x^2y$

Which rearranged looks like: $\frac{dy}{y}=3x^2dx$

At this point, in order to solve for y, we need to take the anti-derivative of both sides: $\int \frac{dy}{y}=\int 3x^2dx$

Which equals:

And since this an anti-derivative with no bounds, we need to include the general constant C

So, solving for y, we raise e to the power of both sides:

$e^{ln(y)}=e^{x^3+C}$ which, simplified gives us our answer:

$y=Ce^{x^3}$

Solve the following separable differential equation: with .

$y = 3e^{t^5}$

$y = 5e^{t^5}$

$y = 3e^{t^4}$

$y = 4e^{t^4}$

$y = 3e^{3t^5}$

Explanation

The simplest way to solve a separable differential equation is to rewrite as $\frac{\mathrm{d} y}{\mathrm{d} t}$ and, by an abuse of notation, to "multiply both sides by dt". This yields

Next, we get all the y terms with dy and all the t terms with dt and integrate. Thus,

$\int \frac{dy}{y} = \int 5t^4dt$

Combining the constants of integration and exponentiating, we have

$|y| = e^{t^5 + C}$

$y = \pm e^Ce^{t^5}$

The plus minus and the can be combined into another arbitrary constant, yielding $y = Ce^{t^5}$ .

Plugging in our initial condition, we have

and

$y = 3e^{t^5}$

Solve the following equation

$\frac{1}{2}\frac{dy}{dx} = \sqrt{y+1}\cos(x) \ \ \ y(\pi) = 0$

$y = (\sin{x} + 1)^2 - 1$

$y = (\sin{x} - 1)^2 - 1\textup{}$

$y = (\sin{x} + 1)^2$

$y = (\cos{x} + 1)^2$

$y = (\cos{x} + 1)^2 - 1$

Explanation

This is a separable ODE, so rearranging

$\frac{1}{\sqrt{y+1}} \ dy = 2\cos{x} \ dx$

Integrating

$2(y+1)^{1/2} = 2\sin{x} + C$

Plugging in the initial condition and solving gives us

Solving for gives us

$2(y+1)^{1/2} = 2\sin{x} + 2$

$(y+1) = (\sin{x} + 1)^2$

$y = (\sin{x} + 1)^2 - 1$

Solve the following initial value problem: $6y' = y^4\sin(t)$ , .

$\left(\frac{1 + \cos(t)}{2}\right)^{-1/3}$

$\left(\frac{1 - \sin(t)}{2}\right)^{-1/3}$

$\left(\frac{1 + 2\cos(t)}{2}\right)^{-2/3}$

$\left(\frac{1 - 2\sin(t)}{2}\right)^{-2/3}$

$\left(\frac{\sin(t) - 1}{2}\right)^{1/3}$

Explanation

This is a separable differential equation. The simplest way to solve this is to first rewrite as $\frac{\mathrm{d} y}{\mathrm{d} t}$ and then by an abuse of notation to "multiply both sides by dt." This yields $6dy = y^4\sin(t)dt$ . Then group all the y terms with dy and integrate, getting us to $\int \frac{6}{y^{4}}dy = \int \sin(t)dt = -\frac{2}{y^3} = -\cos(t) + C$ . Solving for y, we have $y = \left(\frac{\cos(t)}{2} + C\right)^{-1/3}$ . Plugging in our condition, we find $1 = \left( \frac{1}{2} + C\right )^{-1/3}$ . Raising both sides to the power of -1/3, we see $1 = \frac{1}{2} + C; C = \frac{1}{2}$ . Thus, our final solution is $y = \left(\frac{1 + \cos(t)}{2}\right)^{-1/3}$

Solve the given differential equation by separation of variables.

$x\frac{dy}{dx}=6y$

Explanation

To solve this differential equation use separation of variables. This means move all terms containing to one side of the equation and all terms containing to the other side.

$x\frac{dy}{dx}=6y$

First, multiply each side by .

Now divide by on both sides.

$\frac{xdy}{y}=6dx$

Next, divide by on both sides.

$\frac{dy}{y}=\frac{6dx}{x}$

From here take the integral of both sides. Remember rules for logarithmic functions as they will be used in this problem.

$\\int\frac{dy}{y}=\int\frac{6dx}{x} \\\ln y=6\ln x \\e^{\ln y}=e^{6\ln x} \\y=e^{\ln x^6} \\y=cx^6$

Is the following differential equation separable? If So, how does the equation separate?

$\frac{dy}{dt}= e^{t+y}$

The differential equation is separable and becomes:

$\frac{dy}{e^y} = e^t dt$

The differential equation is separable and becomes:

$e^y dy = \frac{e^t}{dt}$

This differential equation is autonomous and is therefore not separable.

Explanation

Using exponential rules, we note that $e^{t+y}$ becomes . Meaning that the

differential equation is equivalent to:

$\frac{dy}{dt} = e^t e^y$

which by separation of variables is:

$\frac{dy}{e^y}= e^tdt$

Solve the given differential equation by separation of variables.

$x\frac{dy}{dx}=6y$

Explanation

To solve this differential equation use separation of variables. This means move all terms containing to one side of the equation and all terms containing to the other side.

$x\frac{dy}{dx}=6y$

First, multiply each side by .

Now divide by on both sides.

$\frac{xdy}{y}=6dx$

Next, divide by on both sides.

$\frac{dy}{y}=\frac{6dx}{x}$

From here take the integral of both sides. Remember rules for logarithmic functions as they will be used in this problem.

$\\int\frac{dy}{y}=\int\frac{6dx}{x} \\\ln y=6\ln x \\e^{\ln y}=e^{6\ln x} \\y=e^{\ln x^6} \\y=cx^6$

Is the following differential equation separable, if so, how does the equation separate?

The differential equation is not separable.

The Differential equation is separable and becomes:

Explanation

The differential equation cannot be written as and is therefore not separable.

Separable Variables

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