Differential Equations

Separable Examples

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Use separation of variables to find the general solution to the given differential equation.

0byyt

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Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

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Next do some algebra until all the y terms are on one side and t terms are on the other:

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

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For Campus Learning Assistance Services at UCSB

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

Now we can integrate both sides and solve for y.

dttb

ydy

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For Campus Learning Assistance Services at UCSB

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

Now we can integrate both sides and solve for y.

dttb

ydy

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For Campus Learning Assistance Services at UCSB

Ctlnbyln

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

Now we can integrate both sides and solve for y.

dttb

ydy

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For Campus Learning Assistance Services at UCSB

Ctlnbyln

Ctlnbyln ee

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

Now we can integrate both sides and solve for y.

dttb

ydy

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For Campus Learning Assistance Services at UCSB

Ctlnbyln

Ctlnbyln ee

Ctlnb eey

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

Now we can integrate both sides and solve for y.

dttb

ydy

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Ctlnbyln

Ctlnbyln ee

Ctlnb eey

Ceybtln

This C is really eC.

Since it is still arbitrary we can still call it C.

Use separation of variables to find the general solution to the given differential equation.

0byyt

First, write the derivative in fraction form. 0bydtdy

t

Next do some algebra until all the y terms are on one side and t terms are on the other:

dttb

ydy

dtbydyt

bydtdy

t

0bydtdy

t

typically you want to leave constants on the t side

Now we can integrate both sides and solve for y.

dttb

ydy

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Ctlnbyln

Ctlnbyln ee

Ctlnb eey

Ceybtln

btCy

This C is really eC.

Since it is still arbitrary we can still call it C.

22 yy et2dtdy

yet2yy

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Use separation of variables to find the general solution to the given differential equation.

22 yy et2dtdy

yet2yy

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Use separation of variables to find the general solution to the given differential equation.

dtt2dyey

dtt2e

dyy

2

2

y

y

We’ll need to use a substitution for the y integral:

dudyyy2dydu

yu212

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Use separation of variables to find the general solution to the given differential equation.

22 yy et2dtdy

yet2yy

dtt2dyey

dtt2e

dyy

2

2

y

y

We’ll need to use a substitution for the y integral:

dudyyy2dydu

yu212

22 y21u

21u

21y eeduedyey

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Use separation of variables to find the general solution to the given differential equation.

22 yy et2dtdy

yet2yy

dtt2dyey

dtt2e

dyy

2

2

y

y

We’ll need to use a substitution for the y integral:

dudyyy2dydu

yu212

22 y21u

21u

21y eeduedyey

Cte 2y21 2

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For Campus Learning Assistance Services at UCSB

Use separation of variables to find the general solution to the given differential equation.

22 yy et2dtdy

yet2yy

dtt2dyey

dtt2e

dyy

2

2

y

y

We’ll need to use a substitution for the y integral:

dudyyy2dydu

yu212

22 y21u

21u

21y eeduedyey

Cte 2y21 2

Ct2e 2y2

C is an arbitrary constant, so we can multiply by -2 and still call it C

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Use separation of variables to find the general solution to the given differential equation.

22 yy et2dtdy

yet2yy

dtt2dyey

dtt2e

dyy

2

2

y

y

We’ll need to use a substitution for the y integral:

dudyyy2dydu

yu212

22 y21u

21u

21y eeduedyey

Cte 2y21 2

Ct2e 2y2

C is an arbitrary constant, so we can multiply by -2 and still call it C

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Use separation of variables to find the general solution to the given differential equation.

22 yy et2dtdy

yet2yy

dtt2dyey

dtt2e

dyy

2

2

y

y

)Ct2ln(y

)Ct2ln(y

2

22

Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

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Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

First we separate and integrate

dt)t2tan(y

dy

)t2tan(ydtdy

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Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

First we separate and integrate

dt)t2tan(y

dy

)t2tan(ydtdy

dt)t2sin(2du)t2cos(u

dt)t2cos()t2sin(

dt)t2tan(

We can use a substitution for this integral

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For Campus Learning Assistance Services at UCSB

Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

First we separate and integrate

dt)t2tan(y

dy

)t2tan(ydtdy

dt)t2sin(2du)t2cos(u

dt)t2cos()t2sin(

dt)t2tan(

We can use a substitution for this integral

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For Campus Learning Assistance Services at UCSB

C)t2cos(lnu

dudt

)t2cos()t2sin(

212

1

Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

First we separate and integrate

dt)t2tan(y

dy

)t2tan(ydtdy

We can use a substitution for this integral

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For Campus Learning Assistance Services at UCSB

)t2cos(lnC)t2cos(ln

21

21

21

Ceey

C)t2cos(lnyln

dt)t2sin(2du)t2cos(u

dt)t2cos()t2sin(

dt)t2tan(

C)t2cos(lnu

dudt

)t2cos()t2sin(

212

1

Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

First we separate and integrate

dt)t2tan(y

dy

)t2tan(ydtdy

Now we can find the constant from the initial value:

2CCe2

Ce20

)02cos(ln21

We can use a substitution for this integral

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For Campus Learning Assistance Services at UCSB

)t2cos(lnC)t2cos(ln

21

21

21

Ceey

C)t2cos(lnyln

dt)t2sin(2du)t2cos(u

dt)t2cos()t2sin(

dt)t2tan(

C)t2cos(lnu

dudt

)t2cos()t2sin(

212

1

Find a particular solution to the given initial value problem.

2)0(y);t2tan(yy

First we separate and integrate

dt)t2tan(y

dy

)t2tan(ydtdy

Now we can find the constant from the initial value:

2CCe2

Ce20

)02cos(ln21

Plugging this into our general solution we get

)t2cos(

2e2y

)t2cos(lnp

21

We can use a substitution for this integral

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)t2cos(lnC)t2cos(ln

21

21

21

Ceey

C)t2cos(lnyln

dt)t2sin(2du)t2cos(u

dt)t2cos()t2sin(

dt)t2tan(

C)t2cos(lnu

dudt

)t2cos()t2sin(

212

1

Find a particular solution to the given initial value problem.

4)1(y;y3yt2

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Find a particular solution to the given initial value problem.

4)1(y;y3yt2

t2dt3

ydy

y3dtdy

t2

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Find a particular solution to the given initial value problem.

4)1(y;y3yt2

t2dt3

ydy

y3dtdy

t2

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Ctlnyln23

Find a particular solution to the given initial value problem.

4)1(y;y3yt2

t2dt3

ydy

y3dtdy

t2

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Ctlnyln23

tln23

Cey

Find a particular solution to the given initial value problem.

4)1(y;y3yt2

t2dt3

ydy

y3dtdy

t2

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Ctlnyln23

tln23

Cey

23

Cty

Find a particular solution to the given initial value problem.

4)1(y;y3yt2

t2dt3

ydy

y3dtdy

t2

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Ctlnyln23

tln23

Cey

23

Cty

4C)1(C4 23

Find C from the initial value

Find a particular solution to the given initial value problem.

4)1(y;y3yt2

t2dt3

ydy

y3dtdy

t2

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For Campus Learning Assistance Services at UCSB

Ctlnyln23

tln23

Cey

23

Cty

4C)1(C4 23

23

t4yp

Find C from the initial value

Find a particular solution to the given initial value problem.

3)0(y;1yy)1t( 2

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Find a particular solution to the given initial value problem.

1t

dtdyy

2

3)0(y;1yy)1t( 2

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Find a particular solution to the given initial value problem.

1t

dtdyy

2

3)0(y;1yy)1t( 2

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C)t(tany 1221

Find a particular solution to the given initial value problem.

1t

dtdyy

2

3)0(y;1yy)1t( 2

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C)t(tan2y 12

C)t(tany 1221

Find a particular solution to the given initial value problem.

1t

dtdyy

2

Don’t forget the +/-

3)0(y;1yy)1t( 2

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C)t(tan2y 12

C)t(tany 1221

C)t(tan2y 1

Find a particular solution to the given initial value problem.

1t

dtdyy

2

Don’t forget the +/-

9CC3

C)0(tan23 1

3)0(y;1yy)1t( 2

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C)t(tan2y 12

C)t(tany 1221

C)t(tan2y 1

Find a particular solution to the given initial value problem.

1t

dtdyy

2

Don’t forget the +/-

9CC3

C)0(tan23 1

9)t(tan2y 1p

3)0(y;1yy)1t( 2

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C)t(tan2y 12

C)t(tany 1221

C)t(tan2y 1

Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

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Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

)tln(tdt

ydy

ydtdy

)tln(t

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Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

)tln(tdt

ydy

ydtdy

)tln(t

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Ctlnlnyln

Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

)tln(tdt

ydy

ydtdy

)tln(t

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Ctlnlnyln

tlnCeyCtlnln

Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

)tln(tdt

ydy

ydtdy

)tln(t

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Ctlnlnyln

tlnCeyCtlnln

2lnC)4ln(

Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

)tln(tdt

ydy

ydtdy

)tln(t

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Ctlnlnyln

tlnCeyCtlnln

2lnC)4ln(

2)2ln()2ln(2

)2ln()2ln(

)2ln()4ln(

C2

Find a particular solution to the given initial value problem.

)4ln()2(y;yy)tln(t

)tln(tdt

ydy

ydtdy

)tln(t

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Ctlnlnyln

tlnCeyCtlnln

2lnC)4ln(

2)2ln()2ln(2

)2ln()2ln(

)2ln()4ln(

C2

tln2yp