"How to solve differential equation word problems ?" is a big question having had by all the students who study quantitative aptitude to get prepared for competitive exams.For some students, solving differential equation word problems is never being easy and always it is a challenging one.
The get answer for the question "How to solve differential equation word problems ? "is purely depending upon the question that we have in the topic "Differential Equations". The techniques and methods we apply to solve word problems on differential equations will vary from problem to problem.
The techniques and methods we apply to solve a particular word problem will not work for another word problem in this topic.
though we have different techniques to solve word problems in different
topics of math, let us see the stuff which is needed to solve any differential equation word problems.
To get answer for the question "How
to solve differential equation word problems ? let us look at the steps
involved in solving the differential equation word problem given.
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
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
Step 6 :
Let us use the second and third information in (1).That is
Step 7 :
Plugging C = 85.15 in (1), we get                       A = 85.15 ekt
Our aim is to find the initial weight of the substance.
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 ek(0)
Hence initially there was 85.15 gms (approximately) of the substance.
(Important 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.)
When students look at the steps of the above problem, we hope that the students would have received answer for the question "How to solve differential equation word problems ?".
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