Differential Equations - Initial-Value Problems

Solve for y

None of these answers

Explanation

So this is a separable differential equation. We can think of as $\frac{d^2y}{dx^2}$

and as $\frac{dy}{dx}$ .

Taking the anti-derivative once, we get:

Then using the initial condition

We get that

So Then taking the antiderivative one more time, we get:

and using the initial condition

we get the final answer of:

If is some constant and the initial value of the function, is six, determine the equation.

Explanation

First identify what is known.

The general function is,

The initial value is six in mathematical terms is,

From here, substitute in the initial values into the function and solve for .

$\y=ce^x \6=ce^0 \6=c\cdot 1 \6=c$

Finally, substitute the value found for into the original equation.

$\frac{dy}{dx}=\frac{4x^2-7x}{3y^2+2}$

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+\frac{31}{6}$

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+\frac{5}{6}$

$y^3+2y=\frac{4}{3}x^3+\frac{7}{2}x^2-\frac{31}{6}$

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+C$

$y^3+2y=\frac{4}{3}x^3+\frac{7}{2}x^2+C$

Explanation

So this is a separable differential equation, but it is also subject to an initial condition. This means that you have enough information so that there should not be a constant in the final answer.

You start off by getting all of the like terms on their respective sides, and then taking the anti-derivative. Your pre anti-derivative equation will look like:

$\int3y^2+2dy=\int4x^2-7dx$

Then taking the anti-derivative, you include a C value:

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+C$

Then, using the initial condition given, we can solve for the value of C:

$(1)^3+2(1)=\frac{4}{3}(1)^3-\frac{7}{2}(1)^2+C$

Solving for C, we get

$3=-\frac{13}{6}+C$

$C=\frac{31}{6}$ which gives us a final answer of:

$y^3+2y=\frac{4}{3}x^3-\frac{7}{2}x^2+\frac{31}{6}$

Solve the initial value problem for .

$y = 5e^{10t}$

$y = 10e^{5t}$

$y = 50e^{t}$

$y = e^{10t} + 4$

$y = e^{5t} + 4$

Explanation

We have $\frac{y'}{y} = 10$ so that $\int \frac{y'}{y}dt = \int10dt$ and $\ln|y| = 10t + C$ . Solving for y,

$|y| = e^{10t + C} = e^{C}e^{10t}$ and $y = \pm e^{C}e^{10t} = ce^{10t}$

which we can write because $\pm e^{C}$ is just another arbitrary constant.

Plugging in our initial value, we have $5 = ce^{0} = c$ leaving us with a final answer of $y = 5e^{10t}$ .

Note, this type of equation pops up frequently in the course and is potentially good to just memorize. For , we have $y = by^{at}$

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

With

$y=ln(\frac{e^{2x}}{2}+\frac{e^{4}}{2})$

$y=ln(\frac{e^{2x}}{2}+C)$

$y=Cln(\frac{e^{2x}}{2})$

$y=ln(\frac{e^{2x}}{2}+\frac{3e^{4}}{2})$

$y=e^{2x}+C$

Explanation

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

$e^{y}dy=e^{2x}dx$

Then integrate, and make sure to add a constant at the end

$e^{y}=\frac{1}{2}e^{2x}+C$ To solve for y, take the natural log, ln, of both sides

$y=ln(\frac{1}{}2e^{2x}+C)$ Be careful not to separate this, a log(a+b) can't be separated.

Plug in the initial condition to get:

$4=ln(\frac{1}{2}e^{4}+C)$ So raising e to the power of both sides:

$e^{4}=\frac{1}{2}e^{4}+C$ Solving for C:

$\frac{1}{2}e^{4}=C$ giving us a final answer of:

$y=ln(\frac{e^{2x}}{2}+\frac{e^{4}}{2})$

Solve the separable differential equation

$\frac{dy}{dx}=\frac{sec(y)}{csc(x)}$

with the initial condition $y(\frac{\pi }{4})=\frac{\pi }{4}$

$y=arcsin(-cos(x)+\sqrt{2})$

$y=arcsin(-cos(x)+\frac{\sqrt{2}}{4})$

$y=arcsin(-cos(x)-\frac{\sqrt{2}}{4})$

none of these answers

Explanation

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

Notice that when you divide sec(y) to the other side, it will just be cos(y),

and the csc(x) on the bottom is equal to sin(x) on the top.

Integrating, we get:

so we can plug pi/4 into both x and y:

$\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}+C$ this gives us a C value of $\sqrt{2}$

$sin(y)=-cos(x)+\sqrt{2}$

In order to solve for y, we just need to take the arcsin of both sides:

$y=arcsin(-cos(x)+\sqrt{2})$

Solve the differential equation

$-2y\frac{dy}{dx}+x^2+2x^4=0$

Subject to:

$y=x\sqrt{\frac{x}{3}+\frac{2x^{3}}{5}}$

$y=\sqrt{\frac{x^3}{2}+\frac{2x^{5}}{5}}$

$y=\sqrt{\frac{x^3}{3}+\frac{2x^{5}}{4}}$

$y=\sqrt{\frac{x^3}{2}+\frac{2x^{5}}{5}+C}$

none of these answers

Explanation

So if you rearrange this equation, you will arrive at a separable differential equation by adding the $-2y\frac{dy}{dx}$ to the other side:

$x^2+2x^4=2y\frac{dy}{dx}$ Now, to solve this, multiply the dx to the other side and take the anti-derivative:

$\int x^2+2x^4dx=\int 2ydy$

Then, after the anti-derivative, make sure to add the constant C:

$\frac{x^3}{3}+\frac{2x^5}{5}+C=y^2$ Now, plug in the initial condition that y(0)=0, which will give you a C=0 as well. Then just take the square root, and you arrive at:

$y=\sqrt{\frac{x^3}{3}+\frac{2x^5}{5}}$ Then, to get the correct answer, simplify by factoring out an and pulling it outside of the square root to get: