To solve this differential equation use separation of variables. This means move all terms containing \(\displaystyle y\) to one side of the equation and all terms containing \(\displaystyle x\) to the other side.

\(\displaystyle x\frac{dy}{dx}=6y\)

First, multiply each side by \(\displaystyle dx\).

\(\displaystyle xdy=6ydx\)

Now divide by \(\displaystyle y\) on both sides.

\(\displaystyle \frac{xdy}{y}=6dx\)

Next, divide by \(\displaystyle x\) on both sides.

\(\displaystyle \frac{dy}{y}=\frac{6dx}{x}\)

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

\(\displaystyle \\\int\frac{dy}{y}=\int\frac{6dx}{x} \\\\\ln y=6\ln x \\\\e^{\ln y}=e^{6\ln x} \\\\y=e^{\ln x^6} \\\\y=cx^6\) 

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:  \(\displaystyle \frac{dy}{dx}=3x^2y\)

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

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

Which equals: \(\displaystyle ln(y)=x^3+C\)

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:

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

\(\displaystyle y=Ce^{x^3}\)

The simplest way to solve a separable differential equation is to rewrite \(\displaystyle y'\) as \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} t}\) and, by an abuse of notation, to "multiply both sides by dt". This yields

 \(\displaystyle dy = 5yt^5dt\).

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

\(\displaystyle \int \frac{dy}{y} = \int 5t^4dt\) 

\(\displaystyle ln|y| +K = t^5 +C\)

Combining the constants of integration and exponentiating, we have 

\(\displaystyle |y| = e^{t^5 + C}\)

\(\displaystyle y = \pm e^Ce^{t^5}\)

The plus minus and the \(\displaystyle e^C\) can be combined into another arbitrary constant, yielding \(\displaystyle y = Ce^{t^5}\).

Plugging in our initial condition, we have

\(\displaystyle 3 = C\)

and

\(\displaystyle y = 3e^{t^5}\)

First the differential equation can be separated to:

 \(\displaystyle \frac{dy}{y} = x^8dx\)

And then integrated simply to:

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

Using exponential rules, we note that \(\displaystyle e^{t+y}\) becomes \(\displaystyle e^t e^y\). Meaning that the

differential equation is equivalent to:

 \(\displaystyle \frac{dy}{dt} = e^t e^y\) 

which by separation of variables is:

 \(\displaystyle \frac{dy}{e^y}= e^tdt\) 

The differential equation \(\displaystyle y'= 3cos(ty)\) cannot be written as \(\displaystyle y' = f(t)g(y)\) and is therefore not separable.

To solve this differential equation use separation of variables. This means move all terms containing \(\displaystyle y\) to one side of the equation and all terms containing \(\displaystyle x\) to the other side.

\(\displaystyle x\frac{dy}{dx}=6y\)

First, multiply each side by \(\displaystyle dx\).

\(\displaystyle xdy=6ydx\)

Now divide by \(\displaystyle y\) on both sides.

\(\displaystyle \frac{xdy}{y}=6dx\)

Next, divide by \(\displaystyle x\) on both sides.

\(\displaystyle \frac{dy}{y}=\frac{6dx}{x}\)

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

\(\displaystyle \\\int\frac{dy}{y}=\int\frac{6dx}{x} \\\\\ln y=6\ln x \\\\e^{\ln y}=e^{6\ln x} \\\\y=e^{\ln x^6} \\\\y=cx^6\) 

This is a separable differential equation. The simplest way to solve this is to first rewrite \(\displaystyle y'\) as \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} t}\) and then by an abuse of notation to "multiply both sides by dt." This yields \(\displaystyle 6dy = y^4\sin(t)dt\). Then group all the y terms with dy and integrate, getting us to \(\displaystyle \int \frac{6}{y^{4}}dy = \int \sin(t)dt = -\frac{2}{y^3} = -\cos(t) + C\). Solving for y, we have \(\displaystyle y = \left(\frac{\cos(t)}{2} + C\right)^{-1/3}\). Plugging in our condition, we find \(\displaystyle 1 = \left( \frac{1}{2} + C\right )^{-1/3}\). Raising both sides to the power of -1/3, we see \(\displaystyle 1 = \frac{1}{2} + C; C = \frac{1}{2}\). Thus, our final solution is \(\displaystyle y = \left(\frac{1 + \cos(t)}{2}\right)^{-1/3}\)

This is a separable ODE, so rearranging

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

Integrating

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

Plugging in the initial condition and solving gives us

\(\displaystyle C=2\)

Solving for \(\displaystyle y\) gives us

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

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

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

First, divide by \(\displaystyle x\) on both sides of the equation.

\(\displaystyle \\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\\\frac{dy}{dx}+\frac{3y}{x}=x^3-1\)

Identify the factor \(\displaystyle p(x)\) term.

\(\displaystyle p(x)=\frac{3}{x}\)

Integrate the factor.

\(\displaystyle e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3\)

Substitute this value back in and integrate the equation.

\(\displaystyle \\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\\\x^3y=\int x^6-x^3dx \\\\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c\)

Now divide by \(\displaystyle x^3\) to get the general solution.

\(\displaystyle \\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}\)

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is \(\displaystyle cx^{-3}\).