Question

sin^-1 (dy/dx) =x+y

Ans:- (x+y) ( 1 + tan (x+y/2) ) +2 =0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{dy}{dx} = x + y$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = sin(x + y)$

To solve this,

$\red{\rm :\longmapsto\:Put \: x + y = z}$

$\rm :\longmapsto\:\dfrac{d}{dx}(x + y) = \dfrac{d}{dx}z$

$\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{dz}{dx}$

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{dz}{dx} - 1$

So, on substituting all these values in above, we get

$\rm :\longmapsto\: \dfrac{dz}{dx} - 1 = sinz$

$\rm :\longmapsto\: \dfrac{dz}{dx} = sinz + 1$

On separating the variables, we get

$\rm :\longmapsto\: \dfrac{dz}{1 + sinz} = dx$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{dz}{1 + sinz} =\displaystyle\int\sf dx$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1}{1 + sinz} \times \frac{1 - sinz}{1 - sinz} \: dz =x + c$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1 - sinz}{1 - {sin}^{2} z} \: dz =x + c$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1 - sinz}{{cos}^{2} z} \: dz =x + c$

$\rm :\longmapsto\:\displaystyle\int\sf \bigg[\dfrac{1}{{cos}^{2} z} - \dfrac{sinz}{ {cos}^{2}z }\bigg]\: dz =x + c$

$\rm :\longmapsto\:\displaystyle\int\sf \bigg[ {sec}^{2}z - secz \: tanz\bigg]\: dz =x + c$

$\rm :\longmapsto\:tanz - secz =x + c$

$\rm :\longmapsto\:tan(x + y) - sec(x + y) = x + c$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}} </p><p>$

Given Differential equation is

$\rm :\longmapsto\: {sin}^{ - 1}\dfrac{dy}{dx} = x + y:⟼sin </p><p>−1</p><p> </p><p>dx</p><p>dy</p><p> </p><p> =x+y</p><p>$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = sin(x + y):⟼ </p><p>dx$

$dy</p><p> </p><p> =sin(x+y)</p><p>$

To solve this,

$\red{\rm :\longmapsto\:Put \: x + y = z}:⟼$

$</p><p>\rm :\longmapsto\:\dfrac{d}{dx}(x + y) = \dfrac{d}{dx}z:⟼ </p><p>dx</p><p>d</p><p> </p><p> (x+y)= </p><p>dx</p><p>d</p><p> </p><p> z$

$\rm :\longmapsto\:1 + \dfrac{dy}{dx} = \dfrac{dz}{dx}:⟼1+ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>dx</p><p>dz$

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{dz}{dx} - 1:⟼ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>dx</p><p>dz</p><p> - 1$

So, on substituting all these values in above, we get

$</p><p>\rm :\longmapsto\: \dfrac{dz}{dx} - 1 = sinz:⟼ </p><p>dx$

$dz</p><p> </p><p> −1=sinz$

$\rm :\longmapsto\: \dfrac{dz}{dx} = sinz + 1:⟼ </p><p>dx</p><p>dz</p><p> </p><p> =sinz+1$

On separating the variables, we get

$\rm :\longmapsto\: \dfrac{dz}{1 + sinz} = dx:⟼ </p><p>1+sinz</p><p>dz</p><p> </p><p> =dx</p><p>$

$</p><p>\rm :\longmapsto\:\displaystyle\int\sf \dfrac{dz}{1 + sinz} =\displaystyle\int\sf dx:⟼∫ </p><p>1+sinz</p><p>dz</p><p> </p><p> =∫dx</p><p>$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1}{1 + sinz} \times \frac{1 - sinz}{1 - sinz} \: dz =x + c:⟼∫ </p><p>1+sinz</p><p>1</p><p> </p><p> × </p><p>1−sinz</p><p>1−sinz</p><p> </p><p> dz=x+c$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1 - sinz}{1 - {sin}^{2} z} \: dz =x + c:⟼∫ </p><p>1−sin </p><p>2</p><p> z</p><p>1−sinz</p><p> </p><p> dz=x+c</p><p>$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{1 - sinz}{{cos}^{2} z} \: dz =x + c:⟼∫ </p><p>cos </p><p>2</p><p> z</p><p>1−sinz</p><p> </p><p> dz=x+c$

$\rm :\longmapsto\:\displaystyle\int\sf \bigg[\dfrac{1}{{cos}^{2} z} - \dfrac{sinz}{ {cos}^{2}z }\bigg]\: dz =x + c:⟼∫[ </p><p>cos </p><p>2</p><p> z</p><p>1</p><p> </p><p> − </p><p>cos </p><p>2</p><p> z</p><p>sinz</p><p> </p><p> ]dz=x+c$

$\rm :\longmapsto\:\displaystyle\int\sf \bigg[ {sec}^{2}z - secz \: tanz\bigg]\: dz =x + c:⟼∫[sec </p><p>2</p><p> z−secztanz]dz=x+c</p><p>$

$\rm :\longmapsto\:tanz - secz =x + c:⟼tanz−secz=x+c</p><p>$

$\rm :\longmapsto\:tan(x + y) - sec(x + y) = x + c:⟼tan(x+y)−sec(x+y)=x+c$

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered} </p><p>f(x)$