CBSE Class 11-science Answered

find general solution dy/dx=sec^2x+ytanx,y(0)=0

Asked by pndash1974 | 08 Nov, 2023, 08:40: PM

Expert Answer

begin mathsize 14px style fraction numerator d y over denominator d x end fraction space equals space s e c squared x space plus space y space tan left parenthesis x right parenthesis end style

begin mathsize 14px style fraction numerator d y over denominator d x end fraction space minus space y space tan left parenthesis x right parenthesis space equals space s e c squared x space space end style

Above equation is of the form "

begin mathsize 14px style fraction numerator d y over denominator d x end fraction space plus space y space P left parenthesis x right parenthesis space equals space Q left parenthesis x right parenthesis end style

" that has solution as

begin mathsize 14px style y left parenthesis x right parenthesis space e x p open parentheses integral P space d x close parentheses space space equals space integral space e x p open parentheses integral P d x close parentheses space Q left parenthesis x right parenthesis space d x space plus space C end style

........................... (1)

begin mathsize 14px style e x p space open parentheses integral P space d x close parentheses space space equals space e x p space open parentheses negative integral tan x space d x close parentheses space equals space e x p open parentheses ln open parentheses cos space x close parentheses close parentheses space equals space cos open parentheses x close parentheses end style

begin mathsize 14px style integral space e x p open parentheses integral P d x close parentheses space Q left parenthesis x right parenthesis space d x space equals space integral cos left parenthesis x right parenthesis space s e c squared x space d x space equals space integral s e c left parenthesis x right parenthesis d x space equals space ln open square brackets s e c left parenthesis x right parenthesis plus tan left parenthesis x right parenthesis close square brackets end style

Hence eqn.(1) becomes

begin mathsize 14px style y left parenthesis x right parenthesis space cos left parenthesis x right parenthesis space equals space ln open square brackets s e c left parenthesis x right parenthesis plus tan left parenthesis x right parenthesis close square brackets plus C end style

If we use the initial condition y(0) = 0 in above expression, we get

0 = ln( 1 + 0 ) + C

Hence , we get C = 0

Hence , solution of given differential equation is

begin mathsize 14px style y left parenthesis x right parenthesis space cos left parenthesis x right parenthesis space equals space ln open square brackets s e c left parenthesis x right parenthesis plus tan left parenthesis x right parenthesis close square brackets end style

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Answered by Thiyagarajan K | 08 Nov, 2023, 10:41: PM

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