CBSE Class 11-science Answered
find general solution
dy/dx=sec^2x+ytanx,y(0)=0
Asked by pndash1974 | 08 Nov, 2023, 20:40: PM
![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](https://images.topperlearning.com/topper/tinymce/cache/5ddd288618034d247daa9b0f366a5852.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/ebb15436339b9bc3158b297bd2c14bc4.png)
Above equation is of the form "
" that has solution as
![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](https://images.topperlearning.com/topper/tinymce/cache/b2ca406437b48dd66f2d1c43c85d579e.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/d48a77522c44624e4c35423c46ef2619.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/5797bfc80e3b0d6fbe010e4ec70ff016.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/1a6781ef2e28b9b6252a60e7a7514b97.png)
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](https://images.topperlearning.com/topper/tinymce/cache/7ba6e7639ab18eb9235346b6290970bf.png)
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](https://images.topperlearning.com/topper/tinymce/cache/5639f05892fc4afda90d3dce29f6f8b4.png)
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Answered by Thiyagarajan K | 08 Nov, 2023, 22:41: PM
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