Question

day?
d) Solve : (x-3) /2 + (x -4)/3 = (x-5)/4​

Answer 1

Answer:

where k,m are real constants and k≠0. We can again think of this equation as describing the growth or decay of a quantity x with respect to time t. Now the growth or decay of x is not only at a rate proportional to itself; it also has a constant component m.

We solve this differential equation by a method similar to that used for our first differential equation. In the Appendix, we give an alternative, perhaps more elegant, method.

For simplicity, we consider the case that kx+m is positive. Taking the reciprocal of both sides of the differential equation yields

dtdx=1kx+m,

so that

t=∫1kx+mdx.

Recalling that k and m are just constants, we can perform this integration and obtain

t=1kloge(kx+m)+c,

where c is a constant of integration. Rearranging this equation gives

x=ek(t−c)−mk=1ke−kcekt−mk.

Now the factor 1ke−kc of the first term is a constant, which we call C.

In general (making no simplifying assumptions), any solution is of the form

x(t)=Cekt−mk,

where C is a real constant. You can easily check that any such x is a solution.

Example

Given that x=3 when t=0, find the solution to the differential equation

dxdt=2x−5.

Solution

The general solution of the differential equation is

x(t)=Ce2t+52,

where C is any constant. Substituting t=0, x=3 gives C=12, so the unique solution is

x(t)=12e2t+52.

If k<0 (that is, if x decays), then ekt approaches 0 as t becomes very large, so

limt→∞x=limt→∞(Cekt−mk)=−mk.

Step-by-step explanation:

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