First we must rearrange this separable differential equation so that we can place  $\displaystyle dy$  alone on one side and  $\displaystyle dx$  on the other side with the terms involving  $\displaystyle x$  and any constants. We then integrate each side with respect to the appropriate variable and solve the result for  $\displaystyle y$  to find the general solution of the differential equation:

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

$\displaystyle -3\frac{dy}{dx}=-3x^2+2$

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

$\displaystyle dy=(x^2-\frac{2}{3})dx$

$\displaystyle \int dy=\int \left(x^2-\frac{2}{3}\right)dx$

Remember when integrating, we increase the exponent by one and then divide the whole term by the value of the new exponent. Will we need to integrate each term that contains $\displaystyle x$ in this fashion.

$\displaystyle y=\frac{1}{3}x^3-\frac{2}{3}x+C$

First we must multiply  $\displaystyle dx$  to the right side of the equation so we have the  $\displaystyle y$  terms with  $\displaystyle dy$  and the  $\displaystyle x$  terms with  $\displaystyle dx$.  We can then integrate each side with respect to the appropriate variable and then solve the result for  $\displaystyle y$  to find the general solution for the differential equation:

$\displaystyle y\frac{dy}{dx}=x^2-1$

$\displaystyle (y)dy=(x^2-1)dx$

$\displaystyle \int (y)dy=\int (x^2-1)dx$

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

$\displaystyle \frac{1}{2}y^2=\frac{1}{3}x^3-x+C$

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

$\displaystyle y=\pm \sqrt{\frac{2}{3}x^3-2x+C}$

In order to separate the differential equation such that  $\displaystyle dy$  is with the  $\displaystyle y$  terms and  $\displaystyle dx$  is with the  $\displaystyle x$  terms, we must divide both sides of the equation by  $\displaystyle y$  and then multiply both sides by  $\displaystyle dx$.  We can then integrate each side with respect to the appropriate variable, and then solve for  $\displaystyle y$  to find the general solution to the differential equation:

$\displaystyle \frac{dy}{dx}=2xy$

$\displaystyle \left(\frac{1}{y}\right)dy=(2x)dx$

$\displaystyle \int \left(\frac{1}{y}\right)dy=\int (2x)dx$

$\displaystyle \ln (y)=x^2+C$

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

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

$\displaystyle y=e^Ce^{x^2}$

$\displaystyle y=Ce^{x^2}$

For this separable differential equation, we can see that if we cross multiply we will have the  $\displaystyle x$  term with  $\displaystyle dx$,  and the  $\displaystyle y$  term with  $\displaystyle dy$.  After getting the equation into this form, we can integrate each side with respect to the appropriate variable, and then solve for  $\displaystyle y$  to find the general solution to the differential equation:

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

$\displaystyle (y)dy=(x^3)dx$

$\displaystyle \int (y)dy=\int (x^3)dx$

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

$\displaystyle \frac{1}{2}y^2=\frac{1}{4}x^4+C$

$\displaystyle y^2=\frac{1}{2}x^4+C$

$\displaystyle y=\pm \sqrt{\frac{1}{2}x^4+C}$

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that  $\displaystyle dy$  is on one side with any  $\displaystyle y$  terms and  $\displaystyle dx$  is on the other side with any  $\displaystyle x$  terms. We can then integrate each side with respect to the appropriate variable and solve for  $\displaystyle y$  to find the general solution for the differential equation. Finally, we plug in the given initial condition to determine the value of the constant, which gives us the particular solution:

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

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

$\displaystyle dy=(3x^2-2x)dx$

$\displaystyle \int dy=\int (3x^2-2x)dx$

Remember when integrating, we can use the power rule. This means to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

$\displaystyle y=x^3-x^2+C$

Once we have found the general solution we plug in the initial conditions to solve for C.

$\displaystyle y(0)=2\rightarrow (0)^3-(0)^2+C=2\rightarrow C=2$

$\displaystyle y=x^3-x^2+2$

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that  $\displaystyle dy$  is on one side with any  $\displaystyle y$  terms and  $\displaystyle dx$  is on the other side with any  $\displaystyle x$  terms. For this problem we can see that the desired arrangement is achieved by simply cross multiplying the differential equation. We can then integrate each side with respect to the appropriate variable and solve for  $\displaystyle y$  to find the general solution for the differential equation. Finally, we plug in the  $\displaystyle x$  and  $\displaystyle y$  values of the given point to determine the value of the constant, which gives us the particular solution:

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

$\displaystyle (y)dy=(2x+1)dx$

$\displaystyle \int (y)dy=\int (2x+1)dx$

Remember when integrating, we can use the power rule. The power rule states to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

$\displaystyle \frac{1}{2}y^2=x^2+x+C$

$\displaystyle y^2=2x^2+2x+C$

$\displaystyle y=\pm\sqrt{ 2x^2+2x+C}$

Now that we have the general solution we can plug in the initial values and solve for C.

$\displaystyle y(0)=0\rightarrow \pm \sqrt{2(0)^2+2(0)+C}=0\rightarrow C=0$

$\displaystyle y=\pm\sqrt{ 2x^2+2x}$

This problem requires the use of the product rule.

The product rule tells us that the derivative of,

 $\displaystyle f(x)\cdot g(x)$ is $\displaystyle f'(x)g(x)+f(x)g'(x)$.

In this problem, $\displaystyle f(x)$ is $\displaystyle x^2$, and $\displaystyle g(x)$ is $\displaystyle e^x$.

Therefore,

$\displaystyle f'(x) = 2x$, and $\displaystyle g'(x)$ = $\displaystyle e^x$, so

$\displaystyle f'(x)g(x)+f(x)g'(x) = 2xe^x+x^2e^x$  , and factoring out $\displaystyle xe^x$ gives us $\displaystyle xe^x(x+2)$.

To find the general solution of $\displaystyle y''+y'-6=0$, rewrite this in the form of its characteristic equation. This will be a quadratic.

$\displaystyle r^2+r-6=0$

Factorize and solve for the roots.

$\displaystyle r^2+r-6=(r+3)(r-2)$

$\displaystyle r_1_,_2=-3,2$

Since the roots are distinct, write the specific form of the general solution stated by one of the three homogenous equation rules.

$\displaystyle y=C_1e^r^__1^x+C_2e^r^__2^x$

Substitute the roots in the formula to obtain the general solution.

$\displaystyle y=C_1e^-^3^x+C_2e^2^x$

To solve for the general solution of $\displaystyle y''-9=0$, convert this to its characteristic equation and solve for the roots.

$\displaystyle r^2-9=0$

$\displaystyle r^2-9=(r-3)(r+3)=0$

$\displaystyle r_1_,_2=\pm3$

Write the specific form of the general solution stated by one of the three homogenous equation rules for distinct roots. 

$\displaystyle y=C_1e^r^__1^x+C_2e^r^__2^x$

Substitute the roots to obtain the general solution.

$\displaystyle y=C_1e^-^3^x+C_2e^3^x$

Let $\displaystyle y=x^x$. Find $\displaystyle \frac{dy}{dx}$.

(Hint: Use differential equations.)

Start with $\displaystyle y=x^x$

First, take the natural logarithm of both sides:

$\displaystyle \ln{y}=\ln{(x^x)}$

Simplify:

$\displaystyle \ln{y}=x\ln{(x)}$

Take the derivatives:

$\displaystyle \frac{y'}{y}=\ln{(x)}+1$

Multiply both sides by $\displaystyle y$:

$\displaystyle y'=(\ln{(x)}+1)y$

Substitute the original equation for $\displaystyle y$:

$\displaystyle y'=(\ln{(x)}+1)x^x$