Higher_Mathematics_Part_2
.pdf1.4.17.(4 + x 2 ) y′ − 2xy = x / y .
1.4.18.3y 2 y′ − y3 / x = x + 1.
1.4.19.3xy′ − y = (x2 + 1) y−2 .
1.4.20.2(1 + x2 ) y′ − 2xy = x / y .
1.4.21.2xy′ + y = x2 y−1 .
1.4.22. y′ − |
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y (x + 2) . |
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4 + x2 |
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1.4.23. y′ + |
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y = 0 . |
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x + 1 |
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1.4.24.y′ + 2y = 2 y .
xcos2 x
1.4.25.xy′ + y = y 2 ln x .
1.4.26.3x2 y′ + xy + y−2 = 0 .
1.4.27. xy′ − 2y = 2x3 y .
1.4.28.2y′ − y tg x + y3 tg x = 0 .
1.4.29.(1 − x2 ) y′ − xy = xy2 .
1.4.30.y′ + xy = y 4 (1 − x2 ) .
1.5. Find the general solution and also the particular solution through the point written opposite the equation.
1.5.1. |
y′ sin x = y ln 2 y , y(π / 2) = e . |
1.5.2. |
y′−2y tg x =sec x , y(0) = 1 . |
1.5.3. 2( y − x) = (x + 2y) y′ , y(1) = 0 .
1.5.4. 2y 2 + x 2 − x2 y′ = 0 , y(1) = 0 .
1.5.5. y′(4 + x 2 ) = 4 + y 2 , y(0) = π / 2 .
1.5.6. (x2 + 4) y′ − 2xy = x , |
y(0) = 1 . |
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1.5.7. (xy |
− y) arctg x = x ln x , y(e) = 0 . |
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1.5.8. xy′ = y(1 − ln 2 |
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1.5.9. 2(1 + e x ) yy′ = e x− y2 , |
y(0) = 0 . |
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1.5.10. y′+ y tg x = cos2 x , |
y(π / 4) = 0.5 . |
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1.5.11. y′− y ctg x =sin 2x cos x , |
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y(π / 2) = 0 . |
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1.5.12. y ′+ y tg x = ex cos x , y(0) = |
1 . |
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1.5.13. sin x sin yy′ = cos x cos2 |
y , |
y(π / 2) = 0 . |
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1.5.14. y + x2 + y 2 − xy′ = 0 , |
y(1) |
= 1 . |
1.5.15. |
y′ = 4 + y / x + ( y / x)2 , y(1) = 2 . |
1.5.16. |
y′ sin2 x = y + 1, y(π / 4) = 1. |
1.5.17. |
(x2 + y2 )dy − 2xydx = 0 , y(1) = 2 . |
1.5.18. |
y′ cos2 x = y , y(π / 4) = e . |
1.5.19.2yy ′ = ( y2 −1) ctg x , y(π / 2) = 0 .
1.5.20.tg ydx −x ln xdy = 0 , x(π / 2) = e .
1.5.21. xy′ − y / x = 1/ x , |
y(1) = 0 . |
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1.5.22. y′ = |
y − x |
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y(1) = 0 . |
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1.5.23. |
y′ = |
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1.5.24. |
y′ = |
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− 2y 2 |
y(1) = 1 . |
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y 2 + x2 |
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1.5.25. |
y′ + 3x2 y = 3x5 , |
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= 1 . |
1.5.26. (x + y) y′ = y − 2x , y(1) = 0 . |
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1.5.27. xy′ = y(1 + ln |
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1.5.28. x2 y′ − y 2 = 1 , y(1) = 0 . |
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1.5.29. (x2 + 1) y′x tg y , |
y(0) = π / 6 . |
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1.5.30. yy′ + xey2 = 0 , |
y(1) = 0 . |
1.6. Solve the exact differential equations.
1.6.1.(4x3 y3 + 3x2 y 2 + 2xy)dx + (3x4 y 2 + 2x3 y + x2 )dy = 0 .
1.6.2.(4x3 y 2 + 3x2 y + 2x)dx + (2x4 y + x3 + 2y)dy = 0 .
1.6.3.(ln x + 2xy 2 )dx + (2x2 y + ln y)dy = 0 .
1.6.4.(cos x sin y + xe x )dx + (sin x cos y + ye y )dy = 0 .
1.6.5.(ln x + y)dx +(ln y + x)dy = 0 .
1.6.6.(arctg x +ln y)dx +( y /(1+ y2 ) + x / y)dy = 0 .
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1.6.7. |
(2x sin y + 3x2 )dx + (x2 cos y + 1/ y)dy = 0 . |
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1.6.8. |
(3x2e y + |
x )dx + (x3e y + y3 )dy = 0 . |
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1 + x2 |
1.6.9.( y + sin x cos2 x)dx + (x + cos y sin3 y)dy = 0 .
1.6.10.(2x cos(x2 + y2 ) + x2 )dx +(2y cos(x2 + y2 ) + y)dy = 0 .
1.6.11.(x( y +2)exy +2x)dx +(2x + x2 exy )dy = 0 .
1.6.12.( y cos x +cos y)dx +(sin x −x sin y +2y)dy = 0 .
1.6.13.(2x sin y +4x3 )dx +(x2 cos y −sin y)dy = 0 .
1.6.14. |
(y −x |
1+ x2 )dx +(y |
y2 −1 + x)dy = 0 . |
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1.6.15. |
(2xe x2 + y + cos x)dx + (e x2 + y − sin y)dy = 0 . |
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1.6.16. |
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+2xy dx + |
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dy = 0 . |
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1.6.17. |
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cos(ln x) + ln y |
dx + |
ln x − sin(ln y) |
dy = 0 . |
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1.6.18.(ex−y + y2 +3x2 )dx +(2xy −ex−y )dy = 0 .
1.6.19.(2xex2 +y2 +3x2 )dx +(2 yex2 +y2 −3y2 )dy = 0 .
1.6.20.(4x3 y2 +2xy3 )dx +(2yx4 +3x2 y2 +4y3 )dy = 0 .
1.6.21.(3x2 y + y2 +2x)dx +(x3 +2xy)dy = 0 .
1.6.22.(sin x + y)dx + ( y cos y2 + x)dy = 0 .
1.6.23.(ln x + ex+ y )dx + (ex + ey )ey dy = 0 .
1.6.24.(2xey + y3ex +2)dx +(x2 ey +3y2 ex )dy = 0 .
1.6.25.(x−1 +2xy2 )dx +( y−1 +2x2 y)dy = 0 .
1.6.26.(4x3 sin y + 2x cos y)dx + (x4 cos y − x2 sin y)dy = 0 .
1.6.27.(y 2 + 3x 2 y 4 + 2x)dx + (2xy + 4x3 y3 − 3y 2 )dy = 0 .
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1.6.28. |
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+ln y +2x dx + ln x + |
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+2y dy = 0 . |
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1.6.29. |
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dy = 0 . |
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+ y dx + x − |
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1.6.30. |
(sin2 x +2xy2 )dx +(2x2 y −cos2 y)dy = 0 . |
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204 |
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And it appears that we may continue this process until we find y in terms of x by n successive integrations:
y = ∫…∫ |
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x n−1 |
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f (x)dx…dx + C1 |
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+ …+ Cn−1 x + Cn . |
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(n − 1)! |
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n |
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Various types of differential equations with appropriate substitution will be considered in the following articles (see table 3.1).
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Table 3.1 |
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Equation of the |
Characteris- |
Substitu- |
The second |
Equation of the |
second order |
tics of |
tion |
derivative |
first order |
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equations |
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Dependent |
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F(x, y′, y′′) = 0 |
variable y |
y ′ = z(x) |
y ′′ = z ′ |
F(x, z, z′) = 0 |
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absent |
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Independent |
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F ( y, p, p′p) = 0 |
F( y, y′, y′′) = 0 |
variable x |
y ′ = p( y) |
y′′ = p′p |
or |
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absent |
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Ф( y, p, p ′) = 0 |
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Consider other types of differential equations with appropriate substitution for reduction of order:
1) a differential equation
F(x, y(k) , y(k +1) , ..., y(n) ) = 0,
which does not contain y directly and derivatives of order less than ( k − 1 ) using the substitution y(k) = z(x) . After substituting, we have
F(x, z, z′, ..., z (n−k) ) = 0 ,
where order is ( n − k );
2) a differential equation
F( y, y′, ..., y(n) ) = 0,
which does not contain independent variable x . It can be a lesser order of times one with substituting the new function:
y′ = p(y),
where p(y) is a new function of y. Thereafter
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dp( y) |
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dp dy |
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d( p′p) |
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d( p′p) |
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dy |
′′ |
′ 2 |
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y |
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= dy dx |
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= ( p p + ( p ) |
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It can be expressed that the order of times one is less as well. |
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Consequently, |
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Ф( y, |
p, |
p′, ..., |
p(n−1) ) = 0 . |
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206 |
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2.3. Linear differential equations of the order higher than the first
The linear differential equation of the n’th order
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a (x) y(n) +a (x) y(n−1) |
+...+a |
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(x) y = f (x), |
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( 3.20 ) |
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0 |
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where |
a0 (x), a1 (x), ..., |
an (x), f (x) |
are the |
given functions, |
moreover |
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≠ 0 is called linear differential equation of order n, since each term is of |
the first degree in the form of y and contains its derivatives, but does not include their products.
An equation (3.20) if f (x) ≠ 0 is called right-hand member not zero. If f (x) ≡ 0 as shown,
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(x) y(n) + a (x) y(n−1) + ... + a |
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(x) y = 0 |
(3.21 ) |
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it’s called homogeneous. |
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Properties of the linear homogeneous differential equation |
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1. If |
y1 is a particular solution |
of (3.21) then C1 y1 ( C1 = const) is a |
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particular solution of (3.21) too. |
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2. If |
y1 , y2 are the particular solutions of (3.21) then the sum of |
y1 + y2 |
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and a linear combination C1 y1 + C2 y2 |
are the solutions of (3.21). |
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In general case: if |
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y1 , y2 , ..., yn are the particular solutions of (3.21) then |
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we are having linear combination |
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y = C1 y1 + C2 y2 + ... + Cn yn
is the solution of (3.21).
There are linear dependent and independent systems of the functions.
System of |
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functions y1 (x), y2 (x), ..., yn (x) |
is called a |
linear |
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dependent on the range (a, b), if an identity |
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α1 y1 + α 2 y2 + ... + α n yn ≡ 0 , |
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(3.22) |
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where α1 , α 2 , ..., α n are real numbers, exists if and only if |
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α1 =α2 =... =αn |
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If at least one of the terms among |
α1 , α 2 , ..., α n is not equal to zero and |
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the identity (3.22) |
exists, then |
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functions y1 , y2 , ..., yn |
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linear |
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independent |
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For instance |
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y1 = sin2 x, |
y2 = cos2 x, |
y3 = 1 |
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linearly |
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dependent, when |
α1 = 1, α 2 = 1, α 3 = −1 exists an identity |
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sin 2 x + cos2 x −1 ≡ 0, |
x (−∞; ∞). |
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2. |
Solve |
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y′′(1+ x2 ) = 2xy′. |
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Solution. The given equation contains derivatives of the dependent variable |
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y but does not contain y directly, then the substitution |
y′ = z(x) (see table 3.1). |
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Then |
y′′ = z′ and we obtain |
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z′(1 + x2 ) = 2xz. |
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Separating the variables in the equation and integrating, we have |
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dz |
(1+ x2 ) = 2xz , |
dz |
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dx , |
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dx |
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+ x2 |
dy |
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ln | z | = ln(1+ x2 ) + ln | C | , z =C (1+x2 ) , |
= C (1+ x2 ) , |
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y = ∫C1 (1 + x2 )dx = C1 (x + |
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is general solution of the given equation. |
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3. |
Solve |
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(y′)2 + 2yy′′ = 0. |
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(3.23) |
Solution. An equation does not contain x , directly, the substitution
y′ = p( y) , y′′ = p dpdy
will give a new differential equation in p and y of order lesser than the original equation
p2 + 2yp |
dp |
= 0; |
p( p + 2y |
dp |
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dy |
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thereafter, the following two cases: 1) p = 0 , dydx = 0 , y =C ;
2) p +2y |
dp |
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dp |
= 0 , |
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ln |
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= ln C , |
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y p =C1 , y |
dy |
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= C1 , ∫ |
ydy = ∫ C1dx , |
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= C1 x + C2 – |
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3 |
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is general solution of the original equation (3.23).
Particular, that solution y =C is proceeded from the general solution, if C1 = 0 .
4. Solve
y′′′−(y′′)2 =0 .
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Solution. An equation is of order 3 and does not contain x and y, directly,
and it depends on y′. |
The substitution |
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necessary, as follows |
y ′′(x) = z(x) or y′ = p( y) . |
It’s suitable to use |
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substitution y ′′(x) = z(x) |
.Eventually we would obtain the resultant equation in the form of First order, as follows; z ′−z2 = 0 .
Then we have:
dz |
= z2 , |
dz |
= dx , − |
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= x +C , z = |
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, y′′ = |
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dx |
z2 |
z |
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−x −C1 |
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−x −C1 |
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y′ = ∫ |
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dx =−ln |
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x +C1 |
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+C2 ; |
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y = ∫ (−ln x +C1 +C2 )dx =−(x +C1 ) ln x +C1 +C2 x +C3 –
Is the general solution of the given equation ( C1 , C2 , C3 –constants)
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Т.2 |
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Self-tests and class assignments |
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Evaluate the following equations: |
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1. |
y′′′ = |
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2. y′′′ = 27e3x + 120x3 . |
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3. |
y′′′ = e5x − cos x . |
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4. |
y′′′ = sin 2x + cos x . |
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5. |
y′′′ = e x + x . |
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6. |
y′′′ = x2 + cos3x . |
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7.. |
y′′ = ln x + x . |
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8. |
y′′ = 5x + 6x . |
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9. |
y′′′ = 3x + cos x . |
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10. |
y′′′ = 4x + cos 3x + sin 5x . |
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11. |
x |
2 |
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+ xy |
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= 1 . |
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12. |
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x |
+ 1)+ y |
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= 0 . |
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y |
(e |
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13. (1− x2 )y′′ − xy′ = 2 . |
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14. |
y′′ = 5y′ . |
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15. |
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2 |
+ y |
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16. |
y |
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+ |
(y |
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= 2e |
− y |
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y (1+ y) = (y ) |
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17. |
y3 y′′ = −1. |
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18. |
yy′′ = (y′)2 . |
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19. |
y′′ = |
1 . |
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20. |
y′′ = 10e y . |
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Find out the linear dependency of the given system of functions |
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21. |
y |
= 1, y |
2 |
= sin 2 x, y |
3 |
= cos 2x . |
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1 |
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22. |
y |
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= x2 − x + 3, |
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2 |
= 2x2 |
+ x, y |
3 |
= 2x − 4 . |
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23. |
1 |
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y1 = x + 2, |
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= x − 2 . |
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210 |
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