Solve in english

We apply the product rule for the first term $\frac{d}{dx}((x^2-3)\ln y) = (2x)\ln y + (x^2-3)\left(\frac{1}{y}\frac{dy}{dx}\right)$.
The derivative of $bx$ is $b$.
The derivative of 4 is 0.
So, we have:

$$2x \ln y + (x^2-3)\frac{1}{y}\frac{dy}{dx} + b = 0$$

Subtract $2x \ln y$ and $b$ from both sides:

$$(x^2-3)\frac{1}{y}\frac{dy}{dx} = -2x \ln y - b$$

Multiply both sides by $y$:

$$(x^2-3)\frac{dy}{dx} = -y(2x \ln y + b)$$

Divide both sides by $(x^2-3)$:

$$\frac{dy}{dx} = \frac{-y(2x \ln y + b)}{x^2-3}$$

To match the desired form, we can factor out $-1$ from the denominator:

$$\frac{dy}{dx} = \frac{y(2x \ln y + b)}{-(x^2-3)}$$ $$\frac{dy}{dx} = \boxed{\frac{y(2x \ln y + b)}{3-x^2}}$$

) b)
Find the equation of the tangent to the curve at the point $P(2, e^2)$.

Substitute the point $(x, y) = (2, e^2)$ into the original curve equation $(x^2-3)\ln y + bx = 4$:

$$(2^2-3)\ln(e^2) + b(2) = 4$$ $$(4-3)(2) + 2b = 4$$ $$1(2) + 2b = 4$$ $$2 + 2b = 4$$ $$2b = 2$$ $$b = 1$$

Use the derivative formula from part (a) with $x=2$, $y=e^2$, and $b=1$:

$$m = \frac{dy}{dx} \Big|_{(2, e^2)} = \frac{e^2(2(2) \ln(e^2) + 1)}{3-2^2}$$ $$m = \frac{e^2(4(2) + 1)}{3-4}$$ $$m = \frac{e^2(8 + 1)}{-1}$$ $$m = \frac{9e^2}{-1}$$ $$m = -9e^2$$

Using the point-slope form $Y - y_1 = m(X - x_1)$ with $(x_1, y_1) = (2, e^2)$ and $m = -9e^2$:

$$Y - e^2 = -9e^2(X - 2)$$ $$Y - e^2 = -9e^2X + 18e^2$$ $$Y = -9e^2X + 18e^2 + e^2$$ $$Y = \boxed{-9e^2X + 19e^2}$$

Problem 5ii)
Solve the differential equation $(1+x^2)\frac{dy}{dx} = 2xy$, given that $y=1$ when $x=0$.

Divide both sides by $y(1+x^2)$:

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

Multiply by $dx$:

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

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

The left side integrates to $\ln|y|$.
For the right side, let $u = 1+x^2$, so $du = 2x\,dx$. The integral becomes $\int \frac{1}{u} du = \ln|u| = \ln(1+x^2)$ (since $1+x^2$ is always positive).
So, we have:

$$\ln|y| = \ln(1+x^2) + C$$

Given $y=1$ when $x=0$:

$$\ln|1| = \ln(1+0^2) + C$$ $$0 = \ln(1) + C$$ $$0 = 0 + C$$ $$C = 0$$

Substitute $C=0$ back into the general solution:

$$\ln|y| = \ln(1+x^2)$$

Exponentiate both sides to solve for $y$:

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

Since $y=1$ (positive) when $x=0$, and $1+x^2$ is always positive, we can remove the absolute value:

$$y = \boxed{1+x^2}$$