Find the particular solution of the differential equation dx dy + x y=2y + y^ 2 y, y 0 given that x = 0 when y= 2

We have, dx dy + x y=2y + y^ 2 y ...(1) Clearly, it is a linear differential equation of the form dx dy +Px=Q Where P= and Q=2y + y^ 2 y I.F.=e^ =e^ dy =e^ | | = Multiplying both sides of (1) by I.F.= , we get ( dx dy +x )= (y^2 + 2y ) y dx dy +x =y^2 +2y Integrating both sides with respect to y, we get x y= ^ 2 dy+ 2y dy+C =y^2 dy- [ d dy (y^2) dy ]dy+ 2y dy+C =y^2 - 2y + 2y dy + C y=y^2 y + C Now, 0 2 = ^2 4 2 +C =- ^2 4 Putting the value of C, we get x y=y^2 y- ^2 4 Hence, x y=y^2 y- ^2 4 is the required solution.

Find the particular solution of the differential equation dx dy +…

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Question

Find the particular solution of the differential equation $\frac{\text{dx}}{\text{dy}} + \text{x}\cot \text{y}=2\text{y} + \text{y}^{2} \cot \text{y},\text{ y}\neq0$ given that x = 0 when $\text{y}=\frac{\pi}{2}$

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Answer

We have,
$\frac{\text{dx}}{\text{dy}} + \text{x}\cot \text{y}=2\text{y} + \text{y}^{2} \cot \text{y}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dx}}{\text{dy}}+\text{Px}=\text{Q}$
Where $\text{P}=\cot\text{y}$ and $\text{Q}=2\text{y} + \text{y}^{2} \cot \text{y}$
$\therefore\ \text{I}.\text{F}.=\text{e}^{\int\text{P}\text{dy}}$
$=\text{e}^{\int\cot\text{y}\text{ dy}}$
$=\text{e}^{\log|\sin\text{y}|}$
$=\sin\text{y}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{y},$ we get
$\sin\text{y}\Big(\frac{\text{dx}}{\text{dy}}+\text{x}\cot\text{y}\Big)=\sin\text{y}\big(\text{y}^2\cot\text{y} + 2\text{y}\big)$
$\Rightarrow\sin \text{y}\frac{\text{dx}}{\text{dy}}+\text{x}\cos\text{y}=\text{y}^2\cos\text{y}+2\text{y}\sin\text{y}$
Integrating both sides with respect to y, we get
$\text{x}\sin \text{y}=\int\text{y}^{2}\cos\text{y}\text{ dy}+\int2\text{y}\sin\text{y}\text{ dy}+\text{C}$
$\Rightarrow\text{x}\sin\text{y}=\text{y}^2\int\cos\text{y dy}-\int\Big[\frac{\text{d}}{\text{dy}}(\text{y}^2)\int\cos\text{y dy}\Big]\text{dy}+\int2\text{y}\sin\text{y dy}+\text{C}$
$ \Rightarrow\text{x} \sin\text{y}=\text{y}^2 \sin\text{y}-\int2\text{y}\sin\text{y} + \int2\text{y}\sin\text{y}\text{ dy} + \text{C}$
$\Rightarrow\text{x}\sin \ \text{y}=\text{y}^2 \sin \ \text{y} + \text{C}$
Now,
$\therefore 0\times\sin \frac{\pi}{2}=\frac{\pi^2}{4} \sin \frac{\pi}{2}+\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{4}$
Putting the value of C, we get
$\text{x}\sin \ \text{y}=\text{y}^2\sin \text{y}-\frac{\pi^2}{4}$
Hence, $\text{x}\sin \ \text{y}=\text{y}^2\sin \text{y}-\frac{\pi^2}{4}$ is the required solution.