If y= x x+2 , show that x dy dx =(1-y)y - Vidyadip

Question

If $\text{y}=\frac{\text{x}}{\text{x}+2},$ show that $\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$

✓

Answer

We have, $\text{y}=\frac{\text{x}}{\text{x}+2}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{x}+2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+2)\frac{\text{d}}{\text{dx}}(\text{x})-\text{x}\frac{\text{d}}{\text{dx}}(\text{x}+2)}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2-\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2}{(\text{x}+2)^2}-\frac{\text{x}}{(\text{x}+2)^2}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+2}-\frac{\text{xy}^2}{\text{x}^2} \Big[\because\ \text{x}+2=\frac{\text{x}}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}-\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\text{y}(1-\text{y})$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=(1-\text{y})\text{y}$
Hence, proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Differentiation→Differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Form the differential equation of the family of curves represented by the equation (a being the perimeter):$(2\text{x}+\text{a})^2+\text{y}^2=\text{a}^2$

Find the vector and cartesian equation of the line through the point (5, 2, -4) and which is parallel to the vector $3\hat{\text{i}}+2\hat{\text{j}}-8\hat{\text{k}}.$

Evaluate the following integrals:$\int\frac{\text{x}}{\text{x}^4-\text{x}^2+1}\text{dx}$

If lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} \text{and} \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then find the value of k and hence find the equation of the plane containing these lines.

If $a, b, c$ are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either $a + b + c = 0$ or $a = b= c$.

Find the image of the point with position vector $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ in the plane $\vec{\text{r}}. (2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=4.$ Also, find the position vectors of the foot of the prependicular and the equation of the perpendicular line through $3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$

Evaluate the following integrals:$\int\text{e}^{\text{x}}.\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\text{dx}$

The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that,

At least one of the events will occur,
None of the events will occur.

If $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix},$ show that AB = A and BA = B.

Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$