To find the equilibrium values, substitute \(\displaystyle y'=0\) and solve for \(\displaystyle y\).  The equilibrium values are the solutions of the differential equation in constant form.

\(\displaystyle y^2 + y'=y\)

\(\displaystyle y^2 + 0=y\)

\(\displaystyle y^2-y=0\)

\(\displaystyle y(y-1)=0\)

\(\displaystyle y=0,1\)

Rewrite \(\displaystyle y'= \frac{x^2}{y^\frac{1}{2}}\) so that the same variables \(\displaystyle \textup{x, dx, y, dy}\) are aligned correctly on the left and right of the equal sign.  

\(\displaystyle y^\frac{1}{2}\cdot y'= x^2\)

\(\displaystyle y^\frac{1}{2}\cdot \frac{dy}{dx}= x^2\)

\(\displaystyle y^\frac{1}{2}\cdot {dy}= x^2 \cdot dx\)

Integrate both sides and solve for \(\displaystyle y\).

\(\displaystyle \int y^\frac{1}{2}\cdot {dy}= \int x^2 \cdot dx\)

\(\displaystyle \frac{2}{3}y^\frac{3}{2}=\frac{1}{3}x^3+C\)

\(\displaystyle y^\frac{3}{2}=\left(\frac{3}{2}\right)\left(\frac{1}{3}x^3+C\right)\)

\(\displaystyle y^\frac{3}{2}=\frac{1}{2}x^3+\frac{3}{2}C\)

\(\displaystyle y=\left(\frac{1}{2}x^3+C\right)^\frac{2}{3}\)

This question requires us to use differential equations. Begin as follows:

\(\displaystyle \frac{dy}{dx}=-4x^3+5x\)

\(\displaystyle dy}=-4x^3+5xdx\)

\(\displaystyle y=\int -4x^3+5xdx\)

\(\displaystyle y= -x^4+\frac{5}{2}x^2+c\)

Now, we know that y must pass through the point (4,5), so we can use this point to find c

\(\displaystyle 5= -(4)^4+\frac{5}{2}4^2+c\)

\(\displaystyle 5= -256+40+c\)

\(\displaystyle c=221\)

So our function is as follows:

\(\displaystyle y= -x^4+\frac{5}{2}x^2+221\)

Multiply \(\displaystyle dx\) and divide \(\displaystyle y\) on both sides of the equation \(\displaystyle \frac{dy}{dx}=10y\).

\(\displaystyle \frac{1}{y}dy=10 \: dx\)

Integrate both sides.

\(\displaystyle \int \frac{1}{y}dy=\int 10\: dx\)

\(\displaystyle ln\left |y \right |=10x+C\)

Use base \(\displaystyle e\) to eliminate the natural log.

\(\displaystyle e^{ln(y)}=e^1^0^x^+^C\)

\(\displaystyle y=Ce^1^0^x\)

Remember that the derivative of \(\displaystyle e^u = e^u \cdot u'\).

\(\displaystyle y'=2e^{2x}\).

Now plug in the \(\displaystyle x=0,1\) to find the corresponding values.

\(\displaystyle y'(1)=14.78\)

\(\displaystyle y'(0)=2.\)

Substitute these into the desired formula.

\(\displaystyle y'(1)-y'(0).\)

\(\displaystyle 14.78-2=12.78\)

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 \(\displaystyle x\). 

\(\displaystyle y=x^3-x^2+C\)