Grade 12th passDifferential Calculus

Solution deferential equation dy/dx =cos(x+y)+sin(x+y), is

shivam Prajapati

8 Years agoGrade 12th pass

2 Answers

Arun

8 Years ago

This is the solution. $ln(1+tan\frac{x+y}{2})= x+c$ $ln(1+tan\frac{v}{2})= x+c$ $\int \frac{sec^2\frac{v}{2}dv}{2(1+tan\frac{v}{2})}= \int dx$ $\int \frac{dv}{1+\frac{1-tan^2\frac{v}{2}}{1+tan^2\frac{v}{2}}+\frac{2tan\frac{v}{2}}{1+tan^2\frac{v}{2}}}= \int dx$ $\int \frac{dv}{1+cosv+sinv}= \int dx$ Integrate both sides $\frac{dv}{1+cosv+sinv}= dx$ Using variable and separable method $\frac{dv}{dx}-1= sin(v)+cos(v)$ $\frac{dy}{dx} = sin(x+y)+cos(x+y)$ Put x+y = v1+dy/dx = dv/dxdy/dx = dv/dx – 1

Arun

Approved Tutor Answer8 Years ago

DE : dy/dx = sin (x+y) + cos (x+y).
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Let : u = x + y.

∴ y = u - x

∴ dy/dx = (du/dx) - 1.
_____________________________

∴ the above DE now becomes :

... (du/dx) - 1 = sin u + cos u

∴ du/dx = ( 1 + cos u ) + sin u

∴ du/dx = ( 2 cos² u/2 ) + ( 2 sin u/2. cos u/2 )

∴ dividing both sides by ( 2 cos² u/2 ),

... (1/2) sec² u/2 (du/dx) = 1 + tan u/2

∴ ∫ [ (1/2) sec² u/2 / ( 1 + tan u/2 ) ] du = ∫ dx

∴ ∫ ( 1 / v ) dv = x, ...... v = 1 + tan u/2

∴ ln |v| = x + C

∴ ln | 1 + tan u/2 | = x + C

∴ ln | 1 + tan [(x+y)/2] | = x + C