Question

For the differential equation $xy \frac{dy}{dx}=(x+2)(y+2)$ , find the solution curve passing through the point (1, -1).

Answer 1

Solution:

The given differential equation is

xy dy/dx = (x + 2) (y + 2)

or, {y/(y + 2)} dy = {(x + 2)/x} dx

or, {(y + 2) - 2}/(y + 2) dy = (1 + 2/x) dx

or, {1 - 2/(y + 2)} dy = (1 + 2/x) dx

or, dy - 2 dy/(y + 2) = dx + 2 dx/x

On integration, we get

∫ dy - 2 ∫ dy/(y + 2) = ∫ dx + 2 ∫ dx/x

or, y - 2 log(y + 2) = x + 2 logx + C,

where C is constant of integration

Since the curve passes through (1, - 1),

- 1 - 2 log(- 1 + 2) = 1 + 2 log(1) + C

or, C = - 2

Thus the required curve is

y - 2 log(y + 2) = x + 2 logx - 2

or, y + 2 = x + 2 {logx + log(y + 2)}

or, y + 2 = x + 2 log{x (y + 2)}.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{xy \: \frac{dy}{dx} \: = (x + 2)(y + 2) \: }$

TO DETERMINE

The solution curve passing through the point (1, -1)

CALCULATION

$\displaystyle \sf{xy \: \frac{dy}{dx} \: = (x + 2)(y + 2) \: }$

$\implies \displaystyle \sf{ \frac{y}{y + 2} dy \: = \frac{(x + 2) }{x} \: dx \: }$

$\implies \displaystyle \sf{ \frac{y + 2 - 2}{y + 2} dy \: = (1 + \frac{ 2 }{x}) \: dx \: }$

$\implies \displaystyle \sf{(1 - \frac{ 2}{y + 2} )dy \: = (1 + \frac{ 2 }{x}) \: dx \: }$

On integration

$\displaystyle \sf{ \int \: dy - \int \frac{ 1}{y + 2} dy \: = \int dx +2 \int \frac{1 }{x} \: dx \: }$

$\implies \: \displaystyle \sf{ y - 2 \ln(y + 2) \: = x + 2 \ln x + c\: } \: \: .....(1)$

Where C is integration constant

Now the equation passing through the point ( 1, - 1)

$\: \displaystyle \sf{ - 1 - 2 \ln( - 1 + 2) \: = 1 + 2 \ln 1 + c\: }$

$\implies \: \displaystyle \sf{ - 1 - 2 \ln 1 \: = 1 + 2 \ln 1 + c\: }$

$\implies\displaystyle \sf{ c = - 2 \: }$

Hence the required equation of the curve is

$\displaystyle \sf{ y - 2 \ln(y + 2) \: = x + 2 \ln x - 2\: } \: \:$

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