Question

For each of the differential equation, find a particular solution satisfying the given condition: $cos \bigg\lgroup\frac{dy}{dx}\bigg\rgroup$ =a(a ∈ R); y = 1 when x = 0

Answer 1

For each of the differential equation, find a particular solution satisfying the given condition: $cos \bigg\lgroup\frac{dy}{dx}\bigg\rgroup$ =a(a ∈ R); y = 1 when x = 0

solution : it is given that ,

differential equation : cos(dy/dx) = a, (a ∈ R)

or, dy/dx = cos-¹(a)

or, ∫dy = cos-¹(a)∫dx

or, y = cos-¹(a)x + c

at x = 0, y = 1

so, 1 = cos-¹(a) × 0 + c

or, 1 = 0 + c

or, c = 1

hence, y = cos-¹(a)x + 1

hence, general solution of differential equation is , y = cos-¹(a).x + 1

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The particular solution of the Differential equation

$\displaystyle \sf{ { \cos}^{ - 1} \bigg( \frac{dy}{dx} \bigg) = a }$

satisfying the condition y = 1 when x = 0

CALCULATION

SOLVE USING DEFINITE INTEGRAL

The given differential equation is

$\displaystyle \sf{ { \cos}^{ - 1} \bigg( \frac{dy}{dx} \bigg) = a }$

$\implies \: \displaystyle \sf{ \frac{dy}{dx} = \cos a }$

$\implies \: \displaystyle \sf{ \frac{dy}{dx} = m } \: \: \: \: ( \: \: taking \: \: m \: = \cos a \: )$

$\implies \: \displaystyle \sf{ {dy} = \: m \: {dx} }$

On integration

$\displaystyle \sf{\int\limits_{1}^{y} dy = \displaystyle \: \sf{m} \int\limits_{0}^{ \sf{x}} dx }$

$\implies \displaystyle \sf{\bigg[ \: y \: \bigg]_1^y} = m \: \bigg[ \: x\: \bigg]_0^x$

$\implies \displaystyle \sf{y - 1 = m(x - 0)}$

$\implies \displaystyle \sf{y - 1 = mx }$

$\implies \displaystyle \sf{y - 1 = x \: \: { \cos}^{ - 1} a }$

$\implies \displaystyle \sf{ \frac{y - 1}{x} = \: { \cos}^{ - 1} a }$

$\implies \displaystyle \sf{ \cos\bigg( \frac{y - 1}{x} \bigg)= \: a }$

RESULT

Hence the required particular solution of the Differential equation is

$\boxed{ \displaystyle \sf{ \: \: \: \cos\bigg( \frac{y - 1}{x} \bigg)= \: a \: \: \: }}$

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