Question
For the following differntial equations verify that the accompanying function is a solution:
Differential equation Function
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$ $\text{y}=\frac{\text{a}}{\text{x}+\text{a}}$

Answer

We have $\text{y}=\frac{\text{a}}{\text{x}+\text{a}}$ $\Rightarrow\text{xy}+\text{ay}=\text{a}$ $\Rightarrow\text{xy}=\text{a}(1-\text{y})$ $\Rightarrow\frac{\text{xy}}{1-\text{y}}=\text{a}$ Given differential equation $\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$ Differentiating both sides of (1) with respect to x, we get $\frac{\text{xy}\Big(0-\frac{\text{dy}}{\text{dx}}\Big)-(1-\text{y})\Big(\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big)}{(\text{xy})^2}=0$ $\Rightarrow\text{xy}\Big(-\frac{\text{dy}}{\text{dx}}\Big)-(1-\text{y})\Big(\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big)=0$ $\Rightarrow-\text{xy}\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}+\text{xy}\frac{\text{dy}}{\text{dx}}+\text{y}^2=0$ $\Rightarrow-\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}+\text{y}^2=0$ $\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{y}^2$Hence, the given function is the solution to the given differential equation.

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

Similar questions

Prove that:
$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
A pair of dice is thrown. Let E be the event that the sum is greater than or equal to 10 and F be the event "5 appears on the first-die". Find $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)$. If F is the event "5 appears on at least one die", find $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)$.
The relation S defined on the set R of all real number by the rule aSb iff a ≥ b is:
  1. An equivalence relation.
  2. Reflexive, transitive but not symmetric.
  3. Symmetric, transitive but not reflexive.
  4. Neither transitive nor reflexive but symmetric.
If the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{-2\text{k}}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\text{k}}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{5}$ are perpendicular, find the value of $k$ and, hence find the equation of the plane containing these lines.
$\text{If y}=\text{e}^{\text{a}\cos^{-1}\text{x}},-1\leq\text{x}\leq1,\ \text{show that}(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-\text{a}^2\text{y}=0.$
If $\begin{bmatrix}1&-1&\text{x}\end{bmatrix}\begin{bmatrix}0&1&-1\\2&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0,$ find x.
Using mean value theorem, prove that there is a point on the curve $y = 2x^2 - 5x + 3$ between the points $A(1, 0)$ and $B(2, 1),$ where tangent is parallel to the chord AB. Also, find that point.
Integrate the function in Exercise:$\frac{1}{\sqrt{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}}$
It is given that the Rolle's theorem holds for the function $f(x) = x^3 + bx^2 + cx,$ $\text{x}\in[1,2]$ at the point $\text{x}=\frac{4}{3},$ the values of $b$ and $c.$
Evaluate:$\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1+\sqrt{\tan\text{x}}}$.