09 - Separable Differential Equations.ks-ic - Kuta Software

Kuta Software - Infinite Calculus

Name___________________________________

Separable Differential Equations

Date________________ Period____

Find the general solution of each differential equation. 1)

dy x− y =e dx

2)

1 dy = dx sec 2 y

3)

dy y = xe dx

4)

dy 2 x = dx e 2 y

5)

dy = 2y − 1 dx

6)

dy = 2 yx + yx 2 dx

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-1-

Worksheet by Kuta Software LLC

For each problem, find the particular solution of the differential equation that satisfies the initial condition. You may use a graphing calculator to sketch the solution on the provided graph. dy x− y = 2e , y(1) = ln (2e + 1) dx

7)

2 dy = xy 2 , y(2) = − dx 5

8)

y

−4

9)

−3

−2

y

4

4

3

3

2

2

1

1

−1

1

2

3

4 x

−4

−3

−2

−1

1

−1

−1

−2

−2

−3

−3

−4

−4

dy = 12 x 3 y, y(0) = 2 dx

10)

−3

−2

4 x

y

4

4

3

3

2

2

1

1

−1

3

x dy = − , y(1) = − 2 dx y

y

−4

2

1

2

3

4 x

−4

−3

−2

−1

1

−1

−1

−2

−2

−3

−3

−4

−4

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-2-

2

3

4 x

Worksheet by Kuta Software LLC

Kuta Software - Infinite Calculus

Name___________________________________

Separable Differential Equations

Date________________ Period____

Find the general solution of each differential equation. 1)

dy x− y =e dx y

2)

x

tan y = x + C y = tan −1 ( x + C)

e =e +C x y = ln (e + C)

3)

dy y = xe dx

4)

x2 −e = + C1 2 x2 y = − ln − +C 2

5)

dy = 2y − 1 dx ln 2 y − 1

dy 2 x = dx e 2 y 2y

e 2 = x + C1 2 ln (2 x 2 + C) y= 2

−y

(

1 dy = dx sec 2 y

)

6)

ln y = x 2 +

= x + C1

2 2x Ce + 1 y= 2

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dy = 2 yx + yx 2 dx

x2 +

y = Ce

-1-

x3 + C1 3

x3 3

Worksheet by Kuta Software LLC

For each problem, find the particular solution of the differential equation that satisfies the initial condition. You may use a graphing calculator to sketch the solution on the provided graph. dy x− y = 2e , y(1) = ln (2e + 1) dx

7)

2 dy = xy 2 , y(2) = − dx 5

8)

y

−4

−3

−2

y

y

4

4

3

3

2

2

1

1

−1

1

2

3

4 x

−4

−2

−1

1

−1

−1

−2

−2

−3

−3

−4

−4

dy = 12 x 3 y, y(0) = 2 dx

10)

−3

−2

4 x

y

4

4

3

3

2

2

1

1

−1

3

x dy = − , y(1) = − 2 dx y

y

−4

2

1 x2 1 − = + y 2 2 2 y=− 2 x +1

x

e = 2e + 1 x y = ln (2e + 1)

9)

−3

1

2

3

4 x

−4

−3

−2

−1

1

−1

−1

−2

−2

−3

−3

−4

−4

ln y = 3 x 4 + ln 2 3x4 y = 2e

2

3

4 x

2

y x2 3 =− + 2 2 2 2 y = − −x + 3 , − 3 < x