Differential Equations

xdy - ydx = x³cosxdx
(xdy-ydx)/x² = xcosxdx
d (y/x) = xcosxdx
y/x = ∫xcosxdx = xsinx - ∫sinxdx = xsinx + cosx + c
y = x²sinx + xcosx + cx
y (π) = 0 + π (-1) + cπ = 0 ⇒ c = 1
y = x²sinx + xcosx + x
y (π/2) = (π/2)² (1) + 0 + π/2 = π²/4 + π/2

dy/dx = e^ (3x+4y) = e³? e?
e? dy = e³? dx
∫e? dy = ∫e³? dx
-e? /4 = e³? /3 + C
y (0)=0 ⇒ -1/4 = 1/3 + C ⇒ C = -7/12.
-e? /4 = e³? /3 - 7/12
e? = (7 - 4e³? )/3
y = (-1/4)ln (7-4e³? )/3)
x = -2/3 ln2 = ln (2? ²/³) = ln (1/4¹/³)
e³? = e^ (ln (1/4) = 1/4.
y = (-1/4)ln (7-1)/3) = (-1/4)ln2.
α = -1/4.

Given $\frac{dy}{dx}=\frac{x{y}^2+y}{x}=y^2+\frac{y}{x}$

OR $\frac{dy}{dx}-\frac{y}{x}=y^{2}OR\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{x}.\frac{1}{y}=1.........\left(i\right)$

=> $\frac{x}{y}=-\frac{x^2}{2}+c..........\left(ii\right)$      

Since curve intersect x + 2y = 4 at x = -2 then y = 3 so

From (ii) $\frac{-2}{3}=-2+cORc=2-\frac{2}{3}=\frac{4}{3}$

put x = 3, then $\frac{3}{y}=\frac{-9}{2}+\frac{4}{3}=\frac{-19}{6}$

$\Rightarrow y=\frac{-18}{19}$  

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

  $Y-y=-\frac{dx}{dy}\left(X-x\right)$         

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$

=> (a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  then

3a – 3b – c = 9       .(ii)

& 4a - $2\sqrt[]{2}b$ - c = 12           .(iii)

also given  $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9

$y^2=a\left(x+\frac{\sqrt[]{a}}{2}\right),a>0$

2yy1 = a

$y^2=2y{y}_1\left(x+\frac{\sqrt[]{2y{y}_1}}{2}\right)$

$y=2y_{1}x+y_1\sqrt[]{2y{y}_1}$

${\left(y-2y_{1}x\right)}^2=y_1^2.2y{y}_1=2y{y}_1^3$

$\therefore order=1,degree=3$

Hence, degree – order = 3 – 1 = 2

$e^{siny}cosy\frac{dy}{dx}+e^{siny}cosx=cosx$

Put esin y = t

$e^{siny}cosy\frac{dy}{dx}=\frac{dt}{dx}$

$\therefore \frac{dt}{dx}+tcosx=cosx$

$I.F=e^{{\displaystyle \int cosxdx}}=e^{sinx}$

$\therefore t{e}^{sinx}={\displaystyle \int e^{sinx}cosxdx}$

Put sin x = u, cos xdx = du

$\therefore putx=0,y\left(0\right)=0,1=1+c\Rightarrow c=0$

$\therefore 1+0+0+0=1$

 $\frac{dv}{dt}\propto V\Rightarrow \frac{dv}{dt}=\lambda V\Rightarrow {\displaystyle \underset{1000}{\overset{1200}{\int }}\frac{dv}{v}}=\lambda {\displaystyle \underset{0}{\overset{2}{\int }}dt}\Rightarrow \lambda =\frac{1}{2}\mathcal{l}n\frac{6}{5}$

${\displaystyle \underset{1000}{\overset{2000}{\int }}\frac{dv}{v}}=\frac{1}{2}\mathcal{l}n\left(\frac{6}{5}\right){\displaystyle \underset{0}{\overset{T}{\int }}dt}\Rightarrow T=\frac{2\mathcal{l}n2}{\mathcal{l}n\left(\frac{6}{5}\right)}\Rightarrow k=2ln2$

$\left|f\left(x\right)-f\left(y\right)\right|\le \left|{\left(x-y\right)}^2\right|$

$\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le x-y$

Taking the limit y x on both sides

$\underset{y\to x}{Lt}\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|\le \underset{y\to x}{Lt}\left(x-y\right)$

$\left|f\text{'}\left(x\right)\right|\le 0$

Hence, modulus cannot be zero. Hence f' (x) = 0. Integrating, we get f (x) = c

at x = 0, f (0) = c = 1

$\therefore f\left(x\right)=1>0,\forall x\in R$

Hence, option (A) is correct option.

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$  

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$           

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$       

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$        

$\Rightarrow \frac{dp}{P-900}=0.5dt$

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

  $\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$          

  $\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$

T = 2 ln 18