Question

for the differential equations xy dy/dx=(x+2)(y+2), find the Solution curve passing through the point(1,-1)​

Answer 1

Given:

• Differential equation $\bf{xy\dfrac{dy}{dx}=(x+2)(y+2)}$

• Solution curve passes through point(1,-1)

To Find:

• Solution of the given differential equation

Solution:

We know that,

• $\tt{mln(a)=ln(a^{m}) }$

• $\tt{\displaystyle\int{1dx}=x+c}$

• $\tt{\displaystyle\int{\dfrac{1}{x}dx}=ln(x)+c}$

• $\tt{ln(1)=0}$

• $\tt{ln(a)+ln(b)=ln(ab)}$

Now,

$\longrightarrow\sf{xy\dfrac{dy}{dx}=(x+2)(y+2)}$

$\longrightarrow\sf{\dfrac{y}{y+2}dy=\dfrac{x+2}{x}dx}$

On integrating both the sides, we get

$\longrightarrow\sf{\displaystyle\int{\dfrac{y}{y+2}}dy=\displaystyle\int{\dfrac{x+2}{x}}dx}$

$\longrightarrow\sf{\displaystyle\int{\dfrac{y+2-2}{y+2}}dy=\displaystyle\int{\dfrac{x}{x}dx}+\displaystyle\int{\dfrac{2}{x}}dx}$

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

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

$\longrightarrow\sf{y-2ln(y+2)=x+2ln(x)+c}$

Since, required curve passes through point(1,-1)

So, point(1,-1) will satisfy the  curve

On putting x=1 and y= -1 in equation of required curve, we get

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

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

$\longrightarrow\sf{-1-2(0)=1+2(0)+c}$

$\longrightarrow\sf{-1=1+c}$

$\longrightarrow\sf{c=-2}$

On putting value of c in equation of required curve, we get

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

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

$\longrightarrow\sf{y-x+2=ln(y+2)^{2} +ln(x)^{2} }$

$\longrightarrow\sf{y-x+2=ln(y+2)^{2}(x)^{2}}$

$\longrightarrow\sf{y-x+2=ln\Big((y+2)(x)\Big)^{2}}$

Hence, $\sf\pink{y-x+2=ln\Big((y+2)(x)\Big)^{2}}$ is the solution curve.