Answer to Question #223128 in Differential Equations for Rashinda

Question

The gradient of the tangent to a curve is given by dy/dx = x²+1 / y², y ≠0. The point P(-1,1) lies on the curve .

a)Find the equationof tangent to the cuve at P.

b) Solve the differential equation to find the equation of the curve in the form y= f(x).

Answer 1

a) Let us find the equation $y=y(x_0)+y'(x_0)(x-x_0)$ of tangent to the curve at $P(x_0,y_0),$ where $x_0=-1,\ y_0=1.$ Since $y'(1)=\frac{(-1)^2+1}{1^2}=2,$ it follows that the equation of tangent is $y=1+2(x+1)$ or $y=2x+3.$

b) Let us solve the differential equation $\frac{dy}{dx} = \frac{x^2+1}{ y^2},$ which is equivalent to $y^2dy=(x^2+1)dx$. It follows that $\int y^2dy=\int(x^2+1)dx,$ and hence $\frac{y^3}3=\frac{x^3}{3}+x+C.$ We conclude that $y^3=x^3+3x+C_1,$ and therefore, the general solution of the differential equation is

$y=\sqrt[3]{x^3+3x+C_1}.$