Question

Solve this Differentiate equation –

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ \bf {x}^{2} \dfrac{ {d}^{2}y}{ {dx}^{2} } + 3x \dfrac{dy}{dx} + y = \dfrac{1}{ {(1 - x)}^{2} } \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$


❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​​​​

Answer 1

Answer:

Let the surface charge density of all three spheres be equal to sigma .

Now sphere of radius R has charge Q1 .

Since

Surface charge density = Charge / Surface area

So

\begin{gathered} \sigma = \frac{ Q_1}{4\pi R {}^{2} } \\ \\ \implies \: Q_1 = 4\pi\sigma R {}^{2} \end{gathered}σ=4πR2Q1⟹Q1=4πσR2

Since 4π is constant and sigma is constant for all three spheres so

Q \propto \: R {}^{2}Q∝R2

So we can conclude that

\begin{gathered}Q_1:Q_2:Q_3 \\ = R {}^{2} : (3R ) {}^{2} :(5R) {}^{2} \\ \\ \implies \: Q_1:Q_2:Q_3 \\ = R {}^{2} : 9R {}^{2} :25R {}^{2} \\ = 1 : 9 : 25\end{gathered}Q1:Q2:Q3=R2:(3R)2:(5R)2⟹Q1:Q2:Q3=R2:9R2:25R2=1:9:25

is it true