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 \(\displaystyle cx^{-3}\).

This equation can be put into the form \(\displaystyle y' + p(t)y = q(t)\) as follows:

\(\displaystyle y' - 3t^2y = t^2\). Differential equations in this form can be solved by use of integrating factor. To solve, take \(\displaystyle p(t) = -3t^2\) and solve for \(\displaystyle \mu = e^{\int p(t)dt} = e^{\int -3t^2dt} = e^{-t^3}\)

Note, when using integrating factors, the +C constant is irrelevant as we only need one solution, not infinitely many. Thus, we have set C to 0.

Next, note that \(\displaystyle (\mu y)' = y'\mu + \mu'y = e^{-t^3}(y' - 3t^2y) = \mu q(t) = e^{-t^3}t^2\)

Or more simply, \(\displaystyle (\mu y)' = t^2e^{-t^3}\). Integrating both sides using substitution of variables we find

\(\displaystyle \mu y = \int t^2e^{-t^3}dt = -\frac{e^{-t^3}}{3} + C\)

Finally dividing by \(\displaystyle \mu = e^{-t^3}\), we see

\(\displaystyle y = -\frac{1}{3} + Ce^{t^3}\). Plugging in our initial condition,

\(\displaystyle 0 = -\frac{1}{3} + C\)

So \(\displaystyle C = \frac{1}{3}\)

And \(\displaystyle y = \frac{e^{t^3} - 1}{3}\).

The term \(\displaystyle y^2\) makes the differential equation nonlinear because a linear equation has the form of 

\(\displaystyle y'+f(t)y=g(t)\)

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 \(\displaystyle cx^{-3}\)

For a differential equation to be exact, two things must be true. First, it must take the form \(\displaystyle M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = 0\). In our case, this is true, with \(\displaystyle M = 2xy^2 + y^2 + 1\) and \(\displaystyle N = 2x^2y + 2xy + 3\). The second condition is that \(\displaystyle M_y = N_x\). Taking the partial derivatives, we find that \(\displaystyle M_y =4xy + 2y\) and \(\displaystyle N_x = 4xy + 2y\). As these are equal, we have an exact equation.

Next we find a \(\displaystyle \Psi\) such that \(\displaystyle \Psi_x = M\) and \(\displaystyle \Psi_y = N\). To do this, we can integrate \(\displaystyle M\) with respect to \(\displaystyle x\) or we can integrate \(\displaystyle N\) with respect to \(\displaystyle y.\) Here, we choose arbitrarily to integrate \(\displaystyle M\).

\(\displaystyle \Psi = \int \Psi_x dx = \int Mdx = \int(2xy^2 + y^2 + 1)dx =x^2y^2 + xy^2 + x + h(y)\)

We aren't quite done yet, because when taking a multivariate integral, the constant of integration can now be a function of y instead of just a constant. However, we know that \(\displaystyle \Psi_y = N\), so taking the partial derivative, we find that \(\displaystyle 2x^2y + 2xy + h'(y) = 2x^2y + 2xy + 3\) and thus that \(\displaystyle h'(y) = 3\) and \(\displaystyle h(y)=3y\).

We now know that \(\displaystyle \Psi = x^2y^2 + xy^2 + x + 3y\), and the point of finding psi was so that we could rewrite \(\displaystyle M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = \Psi_x + \Psi_y\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} \Psi}{\mathrm{d} x} = 0\), and because the derivative of psi is 0, we know it must have been a constant. Thus, our final answer is 

\(\displaystyle \Psi = x^2y^2 + xy^2 + x + 3y = c\).

If you have an initial value, you can solve for c and have an implicit solution.

For a differential equation to be exact, two things must be true. First, it must take the form \(\displaystyle M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = 0\). In our case, this is true, with \(\displaystyle M = 2xy^2 + y\) and \(\displaystyle N = x^2y + xy\). The second condition is that \(\displaystyle M_y = N_x\). Taking the partial derivatives, we find that \(\displaystyle M_y =4xy + 1\) and \(\displaystyle N_x = 2xy + y\). As these are unequal, we do not have an exact equation.

Since this is in the form of a linear equation

\(\displaystyle y' + P(x)y = Q(x)\)

we calculate the integration factor

\(\displaystyle I(x) = e^{\int P(x)} = e^{\int-\frac{1}{x} \ dx} = e^{-\ln x} = \frac{1}{x}\)

Multiplying by \(\displaystyle I(x)\) we get

\(\displaystyle (y \times \frac{1}{x})' = e^x\)

Integrating

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

\(\displaystyle y = xe^x + Cx\)

Plugging in the Initial Condition to solve for the Constant we get

\(\displaystyle e - 1 = e + C \implies C = -1\)

Our solution is

\(\displaystyle y = xe^x - x\)

This is a Bernoulli Equation of the form

\(\displaystyle \frac{dy}{dx} + P(x) = Q(x)y^n\)

which requires a substitution

\(\displaystyle v = y^{1-n}\)

to transform it into a linear equation

Rearranging our equation gives us

\(\displaystyle y^{-3} \frac{dy}{dx} - 5y^{-2} = -\frac{5}{2}x\)

Substituting \(\displaystyle v = y^{-2}\)

\(\displaystyle -\frac{1}{2} \frac{dv}{dx} - 5v = -\frac{5}{2}x\)

\(\displaystyle \frac{dv}{dx} + 10v = 5x\)

Solving the linear ODE gives us

\(\displaystyle v = \frac{x}{2} - \frac{1}{20} + Ce^{-10x}\)

Substituting in \(\displaystyle v = y^{-2}\) and solving for \(\displaystyle y\)

\(\displaystyle y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}\)

Rearranging the following equation

\(\displaystyle (2xy^2 + 1)\ dx +2x^2y \ dy = 0\)

This satisfies the test of exactness, so integrating we have

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