Differential equation: Find the equation for tangentline in P(1.3)

Question

Hi guys Im in dire need of help with this one.

A differential equation is given by (dy/dx)+(3x^2)*y=x^2

Define an equation for the tangentline for the graph at P(1.3) the particular solution goes throught the point P(1.3).

Thanks in advance

Answer 1

I assume that by $P(1.3)$ you mean $P(x=1,y=3)$.

Plug $x=1$ and $y=3$ into your ODE and solve for $\frac{dy}{dx}$. $$\frac{dy}{dx}=1^2-3\cdot (1)^2\cdot 3=-8$$

This is your slope at the point $P$

Now use $y=\frac{dy}{dx}|_P\cdot x + y_0=-8x+y_0$ as a general equation for a linear function (tangent).

Plug in $x=1$ and $y=3$ and solve for $y_0=3+8\cdot1=11$.

Your tangent equation is: $y=-8x+11$.