Question

1. Solve the following initial value problem: Dy/DX = cos^2 y/4x-3; y(1)=π/4

2. Let F(X,y) =ycos(x^2 y^2) + y, then

(a) find the first partial derivatives Fx and Fy.

(b) using( 2) (a) above, find dy/dx.

(C) if F (X,y)=0, then find dy/dx using implicit differentiation to confirm your answer in part (2) (b) above.

Answer 1

1.

$\frac{{dy}}{{dx}} = \frac{{{{\cos }^2}y}}{{4x - 3}} \Rightarrow \frac{{dy}}{{{{\cos }^2}y}} = \frac{{dx}}{{4x - 3}} \Rightarrow \frac{{dy}}{{{{\cos }^2}y}} = \frac{1}{4}\frac{{d(4x - 3)}}{{4x - 3}} \Rightarrow tany = \frac{1}{4}\ln (4x - 3) + C \Rightarrow y = arctan\left( {\frac{1}{4}\ln (4x - 3) + C} \right)$

$y(1) = \frac{\pi }{4} \Rightarrow arctan\left( {\frac{1}{4}\ln (4 - 3) + C} \right) = \frac{\pi }{4} \Rightarrow arctanC = \frac{\pi }{4} \Rightarrow C = 1$

Answer: $y = arctan\left( {\frac{1}{4}\ln (4x - 3) + 1} \right)$

2.

$F(x,y) = y\cos \left( {{x^2}{y^2}} \right) + y$

(a) ${{F'}_x} = \frac{{\partial F}}{{\partial x}} = y( - \sin ({x^2}{y^2})) \cdot 2x{y^2} + 0 = - 2x{y^3}\sin ({x^2}{y^2})$

${{F'}_y} = \frac{{\partial F}}{{\partial y}} = \cos \left( {{x^2}{y^2}} \right) + y( - \sin ({x^2}{y^2})) \cdot 2{x^2}y + 1 = \cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin ({x^2}{y^2}) + 1$

Answer: ${{F'}_x} = - 2x{y^3}\sin ({x^2}{y^2})$ , ${{F'}_y} = \cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin ({x^2}{y^2}) + 1$

(b) $\frac{{dy}}{{dx}} = -\frac{{\frac{{\partial F}}{{\partial x}}}}{{\frac{{\partial F}}{{\partial y}}}} = \frac{{ 2x{y^3}\sin ({x^2}{y^2})}}{{\cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin ({x^2}{y^2}) + 1}}$

Answer: $\frac{{dy}}{{dx}} = \frac{{ 2x{y^3}\sin ({x^2}{y^2})}}{{\cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin ({x^2}{y^2}) + 1}}$

(c) ${\left( {y\cos \left( {{x^2}{y^2}} \right) + y} \right)^\prime }_x = 0$

$y'\cos \left( {{x^2}{y^2}} \right) + y( - \sin \left( {{x^2}{y^2}} \right)) \cdot \left( {2x{y^2} + 2{x^2}yy'} \right) + y' = 0$

$y'\left( {\cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin \left( {{x^2}{y^2}} \right) + 1} \right) = 2x{y^3}\sin \left( {{x^2}{y^2}} \right)$

$y' = \frac{{dy}}{{dx}} = \frac{{2x{y^3}\sin \left( {{x^2}{y^2}} \right)}}{{\cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin \left( {{x^2}{y^2}} \right) + 1}}$

Answer: $\frac{{dy}}{{dx}} = \frac{{ 2x{y^3}\sin ({x^2}{y^2})}}{{\cos \left( {{x^2}{y^2}} \right) - 2{x^2}{y^2}\sin ({x^2}{y^2}) + 1}}$