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Au Degree 1st Sem Maths (2017) Question Paper

By Venkat | February 06, 2021

Andhra University Degree 1st Semester Mathematics Question Paper of the year 2017 is available below. Check it now.

Year - 2017
[BS -S 1135/BA -S 1143] 
B.Sc. DEGREE EXAMINATION. 
(Under CBCS) 
First Semester 
Part II — Mathematics 
Paper I - DIFFERENTIAL EQUATIONS 
(Common with B.A./B.Sc.)
(Effective from 2016-2017 admitted batch) 
Time: Three hours        Maximum: 75 marks 
PART A - (5 x 5 = 25 marks) 
Answer any FİVE of the following Eight questions. 
1. Solve (1 + x) dy/dx - xy = 1-x. 
(1 + x) dy/dx - xy = 1-x ను సాధించండి. 
2. Solve (x2 + y2)dx – 2xydy = 0. 
(x2 + y2)dx – 2xydy = 0 ను సాధించండి.
3. Solve y = 2 px - p2 ; p =dy/dx. 
y = 2 px - p2 ; p =dy/dx ను సాధించండి.
4. Solve (D3 +6D2 +12D+8) y = 0, given that for x = 0, y = 1, y'= -2 and y''= 2. 
x = 0 అయినప్పుడు y = 1, y'= -2 మరియు y"= 2 గా ఇచ్చినప్పుడు (D3 +6D2 +12D+8) y = 0 ను సాధించండి.
5. Solve (D2 - 4) y = x2. 
(D2 - 4) y = x2 ను సాధించండి.
6. Solve (D2 + 1) y = xe2x. 
(D2 + 1) y = xe2x ను సాధించండి.
7. Solve (D2 - 2D+5) y = e2xsinx. 
(D2 - 2D+5) y = e2xsinx ను సాధించండి.
8. If y1 = x is a solution of the equation y''-(2/x)y'+(2/x2)y = 0, then find the general solution of  y''-(2/x)y'+(2/x2)y = x logx.
y''-(2/x)y'+(2/x2)y = 0 యొక్క ఒక సాధన  y1 = x అయితే, y''-(2/x)y'+(2/x2)y = x logx యొక్క సామాన్య సాధనమును కనుక్కోండి.
PART B- (5 x 10 = 50 marks) . 
Answer the following (One from each Unit.) 
UNIT I 
9. (a) Solve (y + y2) dx + xy dy = 0. 
(y + y2) dx + xy dy = 0 ను సాధించండి. 
       Or
(b) Solve the dy/dx + xy/1-x2 = x√y.
dy/dx + xy/1-x2 = x√y ను సాధించండి. 
UNIT II
10. (a)  Find the orthogonal trajectories of the family of curves c1x2 + y2 = 1.
c1x2 + y2 = 1 అనేవక్రాల కుటుంబానికి సంబ సంఛేదములను కనుక్కోండి 
       Or
(b) Solve y = yp2+2px. 
y = yp2+2px ను సాధించండి. 
UNIT III 
11. (a) Solve (D2 - 3D + 2) y = e3x, given that y = 0 when x = 0 and x = loge2. 
X = 0 మరియు x = loge2 అయినప్పుడు y= 0 గా ఇస్తే, (D2 - 3D + 2) y = e3x ను సాధించండి. 
       Or
(b) Solve d3y/dx3 + a2 dy/dx = sinax.
d3y/dx3 + a2 dy/dx = sinax ను సాధించండి. 
UNIT IV 
12. (a) Solve d2y/dx2 -6 dy/dx + 13y = 8e3x sin2x and find the general solution. 
d2y/dx2 -6 dy/dx + 13y = 8e3x sin2x ని సాధించి సామాన్య సాధనమును కనుక్కోండి. 
       Or
(b) Find the complete solution of  d2y/dx2 +2 dy/dx + y = xcosx.
d2y/dx2 +2 dy/dx + y = xcosx యొక్క సంపూర్ణ సాధనమును కనుక్కోండి.
UNIT V 
13. (a) Using the variation of parameter method solve the equation y"-2y'+y = ex log x. 
పరామితుల విచరణ పద్ధతిని ఉపయోగించి, y"-2y'+ y = exlog x సమీకరణాన్ని సాధించండి.
       Or
(b) Solve x2 d2y/dx2 -x dy/dx +2y =xlogx.
x2 d2y/dx2 -x dy/dx +2y =xlogx ను సాధించండి.
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  1. Unknown7 April 2021 at 06:08

    2018,2019 papers upload cheyyandi sir

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