Answer to Question #228475 in Calculus for Ntombizethu

Question

A) solve the following initial value problem: dy/dx= cos^2 y/4x-3 ; y(1)=π/4.

B) let F(x,y)=y cos(x^2 y^2)+y, then

(i) find the first partial derivatives F_x and F_y.

(ii) using b(i) above, find dy/dx.

(iii) if F(x,y)=0,then find dy/dx using implicit differentiation to confirm your answer in part (b) (ii) above.

Answer 1

A)

$$\dfrac{dy}{dx}=\dfrac{\cos^2y}{4x-3}$$

$$\dfrac{dy}{\cos^2y}=\dfrac{dx}{4x-3}$$

$$\int\dfrac{dy}{\cos^2y}=\int\dfrac{dx}{4x-3}$$

$$\tan y=\ln|4x-3|+C$$

$$y(1)=π/4, \tan \dfrac{\pi}{4}=\ln|4(1)-3|+C=>C=1$$

$$\tan y=\ln|4x-3|+1$$

B)

$$F(x,y)=y \cos(x^2 y^2)+y$$

(i)

$$F_x=-2xy^3\sin(x^2y^2)$$

$$F_y=\cos(x^2 y^2)-2x^2y^2\sin(x^2y^2)+1$$

(ii)

$$\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}=-\dfrac{-2xy^3\sin(x^2y^2)}{\cos(x^2 y^2)-2x^2y^2\sin(x^2y^2)+1}$$

$$=\dfrac{2xy^3\sin(x^2y^2)}{\cos(x^2 y^2)-2x^2y^2\sin(x^2y^2)+1}$$

(iii)

$$F(x,y)=0=>y \cos(x^2 y^2)+y=0$$

Differentiate both sides with respect to $x$ and use the Chain Rule

$$\dfrac{dy}{dx}\cdot \cos(x^2 y^2)-y\sin(x^2y^2)(2xy^2+2x^2y\cdot\dfrac{dy}{dx})+\dfrac{dy}{dx}=0$$

Solve for $\dfrac{dy}{dx}$

$$\dfrac{dy}{dx}=\dfrac{2xy^3\sin(x^2y^2)}{\cos(x^2 y^2)-2x^2y^2\sin(x^2y^2)+1}$$

The answers are the same.