HOW TO SOLVE DIFFERENTIAL EQUATION WORD PROBLEMS 

When we try to solve word problems on differential equations, in most cases we will have the following equation. 

That is, 

A  =  Ce^kt

In the above equation, we have to find the value of 'k' and 't' using the information given in the question. 

Once we get the value of 'C' and 'k', solving word problems on differential equations will not be a challenging one.

Example :

In a certain chemical reaction the rate of conversion of a substance at time t is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour, 60 grams remain and at the end of 4 hours 21 grams. How many grams of the substance was there
initially ?

Solution:

Step 1 :

Let us understand the given information. There are three information given in the question. 

1. In a certain chemical reaction the rate of conversion of a substance at time "t" is proportional to the quantity of the substance still untransformed at that instant.

2. At the end of one hour, remaining substance is 60 grams. 

3. At the end of 4 hours, remaining substance is 21 grams.

Step 2 :

Target of the question :

How many grams of the substance was there initially?

Step 3 :

Let 'A' be the untransformed substance at any time 't'.

Conversion of the substance 'A' at time 't' can be written using differential coefficient as 'dA/dt'.

Step 4 :

Using the first information we have

dA/dt ∝ A 

dA/dt = kA

dA/A = kdt

Step 5 :

To get 'A', integrate the above equation on both the sides

∫(dA/A) = ∫kdt ----------> log (A) = kt  + c

Taking 'e' as base on both the sides, we have

e^log(A) = e^{kt + c}

A = e^kt.e^c

A = e^kt.C

A = Ce^kt ----(1)

Step 6 :

Let us use the second and third information in (1).

That is,

when t = 1, A = 60 --------> Ce^k = 60 -----(2)

when t = 4, A = 21 -------->Ce^4k = 21 -----(3)

Taking power 4 in (2), we get

C⁴e^4k=60⁴ -----(4)

Dividing (4) by (3), we get

C³ = 60⁴/21

Using log, we have

C = 85.15 (app.)

Step 7 :

Substitute C  =  85.15 in (1).

A = 85.15 e^kt

Our aim is to find the initial weight of the substance.

That is,

the value of 'A' when t  =  0. Because 'A' stands for untransformed substance and  t = 0 is the initial time.

Therefore, to find the no. of grams of the substance initially, we have to plug t = 0 in (1).

When we do so, we get 

A = 85.15 e^k(0)

A = 85.15 e⁰

A = 85.15 (1)

A = 85.15

So, initially there was 85.15 gms (approximately) of the substance.

Note :

In this problem, we have not found the value of 'k'. Because, when we find the initial weight of the substance, we plug t = 0. When t = 0, 'k' vanished and the value of 'k' is not needed.)

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