Задание 1

Найти общее решение дифференциального уравнения:

1. sec²x^.tgy + sec²y^.tgx^.y = 0 (tgx tgy = C)

2. , (у = x sinln|Cx| )

3. y – aу = e^bx, a, b – const,

4. tgy – y xlnx = 0 ( y = arcsinCln|x|)

5. (х – у)dx +(x+y)dy = 0 ( )

6. xy – y = x²cosx (y = xsinx +Cx)

7. lncos y^.dx +xtgy dy= 0 (x = Cln|cosy| )

8. ху = у + х ( +ln|Cx| = 0 )

9. (x+1)y– 3y = (x+1)⁴e^x ( y = (e^x +C)(x+1)³ )

10. 5e^xtgy = y (e^x –1)sec²y ( y=arctgC(e^x – 1)⁵ )

11. xdy =

12. y – y cosecx = tg ( y = tg (x + C) )

13.

14. (y² – xy)dx + x²dy = 0 (ln|Cx| = )

15. y sinx – ycosx = 2xsin²x ( y = (x² + C)sinx )

16.

17. (x + y)y = x ( )

18. (x² +1)y + 4xy = 3

19. (ху² + х) + (у – х²у)у =0 (y² + 1 = C|x² – 1| )

20. (y = –xln|x| + Cx)

21. y + ycosx = 0,5sin2x ( y = sinx – 1 +Ce^–sinx )

22.

23.

24. ycosx – ysinx = sin2x

25. (xy² + y²) – (x²y – x²)y = 0

26. xy – y =

27. y cosx +y sinx = 1 ( y= sinx + Ccosx )

28. (1+y²)(e^2xdx –e^ydy) – (1+y)dy = 0 ( 0,5e^2x – e^y–arctgy –0,5ln(1+y²) = C )

29.

30. y(xcosy + asin2y) = 1 ( x =–2a(siny +1) +Ce^siny )

Задание 2

Найти частное решение дифференциального уравнения:

1. xdy – ylnydx = 0 , x = 1, y = e (у = е^2х)

2. ху = х + у, у(1) = 0 , (у = xln|x| )

3. , y(1) = 3

4. , у(0) = 0 ( )

5. 

6. y – y = , x = 0, y = 1

7. (1+x²)y = 2xy, y(2) = 1 (y = 0,2(x² + 1) )

8. xy – y = ylny – ylnx, у(1) = е² ( y = xe^2x )

9. 3y y² – y³ = x, у(0) = 2

10. x(y² + 1) + y(x² – 1)y = 0 , у(2) = 1

11. , х = е, у = 0,

12. , x = 2, y = 4

13. tgx dy – (y – 1)dx = 0, x = , y = 0 ( y = 1–2sinx )

14. 

15. xdy + (y – xsinx)dx = 0 , y( ) = 0

16. ( )

17. 16. (х – у)dx + xdy = 0, у(1) = 2. (у =2х – хln|x|)

18. y + y – x – 2 = 0 , y(0) =0 (y = e^–x + x – 1)

19. ydy + x e ^y dx = 0, x = 1, y = 0 ( 0,5x² –(y + 1)e^–y = –0,5 )

20. ,

21. ysinx + y cosx = 1 , ( y = sinx + ( – 1)cosx )

22. ,

23. ,

24. y – y = xe^x , y(0) =1 ( y = e^x(0,5x² + 1) )

25. , у(0) =0

26. ,

27. y (2x + 1) = 4x² + 2y – 1, x = 0, y = 1 ( y =(2x+1)(x+C–ln|2x+1|) )

28. 3xdy – ylnydx = 0 , x = 1, y = e

29. , у(1) = 0

30. y +2xy =x , y(0) = 2