76. Which of the following is a homogeneous differential equation:

(A) (4x +6y +5) dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – x³ + y³) dy = 0

(C) (x² + 2y²) dx + (2xy + dy) = 0

(D) y² dx + (x² – xy + y²) dy = 0

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$              

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$  

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

$C=+\frac{\pi }{2}$

now at x = $\mathcal{l}n\sqrt[]{3}$

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$

$x_1=\underset{x\to \infty }{lim}\frac{2^{\frac{x^n}{e^x}}-3^{\frac{x^n}{e^x}}}{\frac{x^n}{e^x}},put\frac{x^n}{e^x}=t$

$x_2=\underset{x\to \infty }{lim}\frac{co{t}^{-1}\left(\sqrt[]{x+1}-\sqrt[]{x}\right)}{se{c}^{-1}\left({\left(\frac{2x+1}{x-1}\right)}^x\right)}$

$x_2=\frac{2}{\pi }\underset{x\to \infty }{lim}ta{n}^{-1}\left(\sqrt[]{x+1}+\sqrt[]{x}\right)$

$x_2=\frac{2}{\pi }.\frac{\pi }{2}=1$

Differentiating

y.  $\frac{-2x}{2\sqrt[]{1-x^2}}+\sqrt[]{1-x^2}\cdot y^{\mathrm{\text{'}}}=\frac{x\cdot 2y{y}^{\mathrm{\text{'}}}}{2\sqrt[]{1-y^2}}-\sqrt[]{1-y^2}$

Put $x=\frac{1}{2},y=-\frac{1}{4}$ and $x,y=-\frac{1}{8}$

$y^{\mathrm{\text{'}}}=\frac{-\sqrt[]{5}}{2}$

dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.