Start by separating variables:

\(\frac{dy}{y^2} = \left(1 + \frac{1}{y^3}\right) dx\)

Integrate both sides:

\(\int \frac{dy}{y^2} = \int \left(1 + \frac{1}{y^3}\right) dx\)

This gives:

\(-\frac{1}{y} = x + C\)

Using the initial condition \(y = 1\) when \(x = 0\), find \(C\):

\(-1 = 0 + C\)

So, \(C = -1\).

The equation becomes:

\(-\frac{1}{y} = x - 1\)

Rearrange to solve for \(y\):

\(y = -\frac{1}{x - 1}\)

Rewriting in terms of \(y^3\), we have:

\(y^3 = (2e^{3x} - 1)\)

Thus, \(y = (2e^{3x} - 1)^{\frac{1}{3}}\).

To solve the differential equation \(\frac{dy}{dx} = \frac{xy}{1+x^2}\), we first separate the variables:

\(\frac{1}{y} \, dy = \frac{x}{1+x^2} \, dx\).

Integrate both sides:

\(\int \frac{1}{y} \, dy = \int \frac{x}{1+x^2} \, dx\).

This gives:

\(\ln |y| = \frac{1}{2} \ln(1+x^2) + C\).

Exponentiate both sides to solve for \(y\):

\(|y| = e^C \sqrt{1+x^2}\).

Using the initial condition \(y = 2\) when \(x = 0\), we find:

\(2 = e^C \sqrt{1+0^2}\), so \(e^C = 2\).

Thus, \(y = 2\sqrt{1+x^2}\).

The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y}{\sqrt{x+1}}\), where \(k\) is a constant of proportionality.

Separate variables: \(\frac{1}{y} dy = \frac{k}{\sqrt{x+1}} dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{k}{\sqrt{x+1}} dx\).

This gives \(\ln y = 2k \sqrt{x+1} + c\), where \(c\) is the constant of integration.

Using the point \((0, 1)\), substitute into the equation: \(\ln 1 = 2k \sqrt{0+1} + c\), which simplifies to \(c = -2k\).

Using the point \((3, e)\), substitute into the equation: \(\ln e = 2k \sqrt{3+1} - 2k\), which simplifies to \(1 = 4k - 2k\).

Solve for \(k\): \(k = \frac{1}{2}\).

Substitute \(k = \frac{1}{2}\) and \(c = -1\) back into the equation: \(\ln y = \sqrt{x+1} - 1\).

Exponentiate both sides to solve for \(y\): \(y = \exp(\sqrt{x+1} - 1)\).

Start by separating the variables:

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

Integrate both sides:

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

This gives:

\(\ln y = \ln x - x^2 + C\)

Use the initial condition \(y = 1\) when \(x = 1\) to find \(C\):

\(\ln 1 = \ln 1 - 1^2 + C\)

\(0 = 0 - 1 + C\)

\(C = 1\)

Substitute \(C\) back into the equation:

\(\ln y = \ln x - x^2 + 1\)

Exponentiate both sides to solve for \(y\):

\(y = e^{\ln x - x^2 + 1}\)

\(y = e^{\ln x} \cdot e^{-x^2} \cdot e^{1}\)

\(y = x \cdot e^{1-x^2}\)

First, separate the variables:

\(\frac{y}{y^2 + 5} dy = \frac{1}{x + 1} dx\)

Integrate both sides:

\(\int \frac{y}{y^2 + 5} dy = \int \frac{1}{x + 1} dx\)

The left side integrates to \(\frac{1}{2} \ln(y^2 + 5)\) and the right side to \(\ln(x + 1)\).

Thus, \(\frac{1}{2} \ln(y^2 + 5) = \ln(x + 1) + C\).

Use the initial condition \(y = 2\) when \(x = 0\) to find \(C\):

\(\frac{1}{2} \ln(4 + 5) = \ln(1) + C\)

\(\frac{1}{2} \ln(9) = C\)

\(C = \frac{1}{2} \ln(9)\)

Substitute back to get:

\(\frac{1}{2} \ln(y^2 + 5) = \ln(x + 1) + \frac{1}{2} \ln(9)\)

Multiply through by 2:

\(\ln(y^2 + 5) = 2\ln(x + 1) + \ln(9)\)

Combine the logarithms:

\(\ln(y^2 + 5) = \ln(9(x + 1)^2)\)

Exponentiate both sides:

\(y^2 + 5 = 9(x + 1)^2\)

Thus, \(y^2 = 9(x + 1)^2 - 5\).