This is a separable differential equation, which means we can separate  $\displaystyle dy$  and  $\displaystyle dx$,  placing each on one side of the equation with its corresponding terms. There are no terms containing  $\displaystyle y$,  so we simply place  $\displaystyle dy$  alone on one side of the equation and  $\displaystyle dx$  on the other side of the equation with any  $\displaystyle x$  terms. We can then integrate each side with respect to the appropriate variable, which gives us an equation for  $\displaystyle y$  that is the general solution to the differential equation:

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

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

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

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

Rearrange the equation to have all the  $\displaystyle C_A$ terms on one side and the rest of the terms on the other side.

$\displaystyle \frac{dC_A}{dt}=-kC_A^2$

$\displaystyle \frac{1}{C_A^2}dC_A=-kdt$

From here we need to take the integral of the function.

$\displaystyle \int \frac{1}{C_A^2}dC_A=\int -kdt$

$\displaystyle -\frac{1}{C_A}=-kt+C$ Where C is some constant.

We know that the initial concentration of A is 10 mols. In otherwords, at

$\displaystyle t=0, C_A=10$

$\displaystyle -\frac{1}{10}=-k(0)+C$

$\displaystyle C=-\frac{1}{10} mols$

Solving for $\displaystyle C_A$ gives

$\displaystyle C_A(t)=\frac{1}{kt+\frac{1}{10}}$.

Plugging everything in gives

$\displaystyle C_A(15)=1.04 mols$.

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:

$\displaystyle x\frac{dy}{dx}-y\sqrt{x}=0$

$\displaystyle x\frac{dy}{dx}=y\sqrt{x}$

$\displaystyle \frac{dy}{dx}=\frac{y}{\sqrt{x}}$

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

$\displaystyle \int \frac{1}{y}dy=\int \frac{1}{\sqrt{x}}dx$

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

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

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

$\displaystyle y(0)=1\rightarrow Ce^{2\sqrt{0}}=1\rightarrow C=1$

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

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  $\displaystyle x$  and  $\displaystyle y$  values of the given point to determine the value of the constant, which gives us the particular solution:

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

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

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

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

$\displaystyle e^{\ln(y)}=e^{-x^{-1}+C}$

$\displaystyle y=e^{-x^{-1}+C}=e^Ce^{-x^{-1}}=Ce^{-\frac{1}{x}}$

$\displaystyle y(1)=-1\rightarrow Ce^{-\frac{1}{1}}=-1\rightarrow C=-e$

$\displaystyle y=(-e)e^{-\frac{1}{x}}=-e^{1-\frac{1}{x}}$

This is a separable differential equation, which means we are able to separate  $\displaystyle dy$  and  $\displaystyle dx$,  placing them on opposite sides of the equation with their corresponding variables. There are no  $\displaystyle y$  terms, so we place  $\displaystyle dy$  alone on one side,  and  $\displaystyle dx$  on the other side with the terms containing  $\displaystyle x$.  We can then integrate each side with respect to the appropriate variable, which gives an equation for  $\displaystyle y$  that is the general solution to the differential equation:

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

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

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

Remember when we integrate, we increase the exponent by one and then divide the term by the value of the new exponent. We will need to integrate each term that contains a $\displaystyle x$. 

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

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.

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

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