Question

If the curve $y=y\left(x\right)$ represented by the solution of the differential equation $(2x{y}^2–y)dx+xdy=0$, passes through the intersection of the lines, $2x–3y=1$ and$3x+2y=8$, then $\left|y\left(1\right)\right|$ is equal to

Answer 1

Step 1: Solve the given differential equation

$(2x{y}^2–y)dx+xdy=0$

$\Rightarrow$ $2x{y}^{2}dx-ydx+xdy=0$

$\Rightarrow$ $2x{y}^{2}dx=ydx-xdy$

$\Rightarrow$ $2xdx=\frac{ydx-xdy}{y^2}$

$\Rightarrow$ $2xdx=d\left(\frac{x}{y}\right)$ $...\left[\because d\left(\frac{x}{y}\right)=\frac{ydx-xdy}{y^2}\right]$…[Quotient rule]

Integrating both sides we get

$\int 2xdx=\int d\left(\frac{x}{y}\right)$

$\Rightarrow$ $x^2=\frac{x}{y}+c$ $...\left(i\right)$ is the equation of the curve

Step 2: Solve the equations of the given lines simultaneously to find the point of intersection

$2x–3y=1$$...\left(ii\right)$

$3x+2y=8$$...\left(iii\right)$

Multiplying $\left(ii\right)$ by $\frac{3}{2}$ we get

$3x–\frac{9}{2}y=\frac{3}{2}$$...\left(iv\right)$

Subtracting equation $\left(iv\right)$ from $\left(iii\right)$ we get

$\frac{13}{2}y=\frac{13}{2}$

$\Rightarrow$ $y=1$

Resubstituting the value of $y$ in $\left(ii\right)$ we get

$2x-3=1$

$\Rightarrow$ $x=2$

Hence, $(2,1)$ is the point of intersection of the given lines.

Step 3: Solve for the required value

Substituting the co-ordinates in equation of the curve we get

$\Rightarrow$$2^2=\frac{2}{1}+c$

$\Rightarrow$ $c=2$

Hence, the equation of the curve is $x^2=\frac{x}{y}+2$

Now to find $\left|y\left(1\right)\right|$ substitute $x=1$ in the equation of the curve

$\Rightarrow$ $1=\frac{1}{y}+2$

$\Rightarrow$ $-1=\frac{1}{y}$

$\Rightarrow$$\left|y\left(1\right)\right|$$=-1$

Hence, the value of $\left|y\left(1\right)\right|$ for the given differential equation and lines is $-1$.