Answer to Question #159524 in Differential Equations for Ojugbele Daniel

Question

Solve the differential equations using separating variables.

1. Cosy + ( 1+ e raise to power of -x ) siny dy/dx equal zero. Given that y= π/4 when x=0.

2. X raise to power of 2 ( y +1 ) + y raise to power of 2 ( x - 1 ) dy/dx equal zero.

Answer 1

Solve the differential equations using separating variables.

1. Cosy + ( 1+ e raise to power of -x ) siny dy/dx equal zero. Given that y= π/4 when x=0.

2. X raise to power of 2 ( y +1 ) + y raise to power of 2 ( x - 1 ) dy/dx equal zero.

Solution:

1. $\cos{y}+(1+e^{-x})\sin{y}\frac{dy}{dx}=0$

$\cos{y}=-(1+e^{-x})\sin{y}\frac{dy}{dx}$

$\frac{dx}{1+e^{-x}}=-\frac{\sin{y}dy}{\cos{y}}$

$\int\frac{dx}{1+e^{-x}}=-\int\frac{\sin{y}dy}{\cos{y}}$

$\int\frac{e^{x}dx}{1+e^{x}}=\int\frac{d\cos{y}}{\cos{y}}$

$\int\frac{d(e^{x}+1)}{1+e^{x}}=\int\frac{d\cos{y}}{\cos{y}}$

$\ln(e^x+1)+\ln{C}=\ln{\cos{y}}$

$\cos{y}=C(e^x+1)$

$\cos{\frac{\pi}{4}}=C(e^0+1)$

$\frac{1}{\sqrt2}=C\cdot2$

$C=\frac{1}{2\sqrt2}=\frac{\sqrt2}{4}$

$\cos{y}=\frac{\sqrt2}{4}(e^x+1)$ .

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

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

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

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

$\int((x-1)+2+\frac{1}{x-1})dx=-\int((y+1)-2+\frac{1}{y+1})dy$

$\int(x+1+\frac{1}{x-1})dx=-\int(y-1+\frac{1}{y+1})dy$

$\frac12(x+1)^2+\ln{|x-1|}=-\frac12(y-1)^2-\ln{|y+1|}+\ln{C}$

$(x-1)(y+1)=Ce^{-\frac12((x+1)^2+(y-1)^2)}$

Answer:

1. $\cos{y}=\frac{\sqrt2}{4}(e^x+1)$

2. $(x-1)(y+1)=Ce^{-\frac12((x+1)^2+(y-1)^2)}$