Example 2 (Particular Solution):

Determine the particular solution to the differential equation dy dx = ( y 1 ) 3 cos ( πx ) , given y (1) = 2 .

Identifier: The directions of the problem include the language “ particular solution ” and “ differential equation ”.

Step 1: Identify your two variables.

In this example the two variables are the y-variable and x-variable .

d y d x = ( y 1 ) 3 cos ( π x )

Step 2: Separate your equation by variables on to either side of the equals.

1) Multiply both sides of the equation by 1 ( y 1 ) 3 to get the y-variables on the same side as d y .

2) Multiply both sides of the equation by dx to get the x-variables on the same side as d x .

d y d x = ( y 1 ) 3 cos ( π x )

1)                1 ( y 1 ) 3 d y d x = ( y 1 ) 3 cos ( π x ) 1 ( y 1 ) 3

2)                dx 1 ( y 1 ) 3 d y d x = cos ( π x ) dx

 

1 ( y 1 ) 3 d y = cos ( π x ) d x

Step 3: Take the antiderivative of both sides of the equation.

Both antiderivatives require you to run a u-substitution process.

Keep in mind that you only need a single + C that works for the entire equation.

1 ( y 1 ) 3 d y = cos ( π x ) d x

1 ( y 1 ) 3 d y

cos ( π x ) d x

1 ( y 1 ) 3 dy

u = y 1

d u d y = 1

d u = d y

1 ( u ) 3 d u

( u ) 3 d u

u 3 d u = u 2 2

u 2 2 = ( y 1 ) 2 2

cos ( πx ) dx

u = πx

d u d x = π

d u = π d x

d u π = d x

cos ( u ) d u π

1 π cos ( u ) d u

1 π cos ( u ) d u = 1 π sin ( u ) + C

1 π sin ( u ) + C = 1 π sin ( πx ) + C

1 ( y 1 ) 3 d y = cos ( π x ) d x

( y 1 ) 2 2 = 1 π sin ( πx ) + C

Step 4: Solve the equation you found in Step 3 to get the dependent variable ( y -variable) alone on one side of the equals.

( y 1 ) 2 2 = 1 π sin ( πx ) + C

2 ( y 1 ) 2 2 = ( 1 π sin ( πx ) + C ) 2

( y 1 ) 2 = 2 π sin ( πx ) + C

( ( y 1 ) 2 ) 1 2 = ( 2 π sin ( πx ) + C ) 1 2

y 1 = ( 2 π sin ( πx ) + C ) 1 2

y 1 + 1 = ( 2 π sin ( πx ) + C ) 1 2 + 1

y = ( 2 π sin ( πx ) + C ) 1 2 + 1

Step 5 (If asked to find a particular solution): Determine the particular +C value for your specific situation.

 

In this example you were given the fact that y ( 1 ) = 2 . Which means that when x = 1 , y = 2 .

1) Plug in 1 for x and 2 for y in the general solution from Step 4, and evaluate any parts of the equation you can. Here you can simplify 2 π sin ( π ( 1 ) ) = 0

2) Subtract one, 1 , from both sides to isolate the part of the equation raised to the 1 2 power.

3) Raise both sides to the 2 power to cancel out the 1 2 power.

4) Plug the C value back into the general solution from Step 4.

y = ( 2 π sin ( πx ) + C ) 1 2 + 1

1)

2 = ( 2 π sin ( π ( 1 ) ) + C ) 1 2 + 1

2 = ( 0 + C ) 1 2 + 1

2 = ( C ) 1 2 + 1

2)

2 1 = ( C ) 1 2 + 1 1  

1 = ( C ) 1 2

3)

( 1 ) 2 = ( ( C ) 1 2 ) 2  

1 = C

4)

y = ( 2 π sin ( πx ) + C ) 1 2 + 1

y = ( 2 π sin ( πx ) + 1 ) 1 2 + 1

Final Result:

The particular solution to the differential equation dy dx = ( y 1 ) 3 cos ( πx ) , given y ( 1 ) = 2 would be y = ( 2 π sin ( πx ) + 1 ) 1 2 + 1 .

 

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