10 simple and solved questions of differential equations
Answers
Answered by
0
Examples of Differential Equations
Example 1
We saw the following example in the Introduction to this chapter. It involves a derivative, \displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}dxdy:
\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={x}^{2}-{3}dxdy=x2−3
As we did before, we will integrate it. This will be a general solution (involving K, a constant of integration).
So we proceed as follows:
\displaystyle{y}=\int{\left({x}^{2}-{3}\right)}{\left.{d}{x}\right.}y=∫(x2−3)dx
and this gives
\displaystyle{y}=\frac{{x}^{3}}{{3}}-{3}{x}+{K}y=3x3−3x+K
But where did that dy go from the \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}dxdy? Why did it seem to disappear?
In this example, we appear to be integrating the xpart only (on the right), but in fact we have integrated with respect to y as well (on the left). DEs are like that - you need to integrate with respect to two (sometimes more) different variables, one at a time.
We could have written our question only using differentials:
dy = (x2 − 3)dx
(All I did was to multiply both sides of the original dy/dx in the question by dx.)
Now we integrate both sides, the left side with respect to y (that's why we use "dy") and the right side with respect to x (that's why we use "dx") :
\displaystyle\int{\left.{d}{y}\right.}=\int{\left({x}^{2}-{3}\right)}{\left.{d}{x}\right.}∫dy=∫(x2−3)dx
Then the answer is the same as before, but this time we have arrived at it considering the dy part more carefully:
\displaystyle{y}=\frac{{x}^{3}}{{3}}-{3}{x}+{K}y=3x3−3x+K
On the left hand side, we have integrated \displaystyle\int{\left.{d}{y}\right.}=\int{1}{\left.{d}{y}\right.}∫dy=∫1dy to give us y.
Example 1
We saw the following example in the Introduction to this chapter. It involves a derivative, \displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}dxdy:
\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={x}^{2}-{3}dxdy=x2−3
As we did before, we will integrate it. This will be a general solution (involving K, a constant of integration).
So we proceed as follows:
\displaystyle{y}=\int{\left({x}^{2}-{3}\right)}{\left.{d}{x}\right.}y=∫(x2−3)dx
and this gives
\displaystyle{y}=\frac{{x}^{3}}{{3}}-{3}{x}+{K}y=3x3−3x+K
But where did that dy go from the \displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}dxdy? Why did it seem to disappear?
In this example, we appear to be integrating the xpart only (on the right), but in fact we have integrated with respect to y as well (on the left). DEs are like that - you need to integrate with respect to two (sometimes more) different variables, one at a time.
We could have written our question only using differentials:
dy = (x2 − 3)dx
(All I did was to multiply both sides of the original dy/dx in the question by dx.)
Now we integrate both sides, the left side with respect to y (that's why we use "dy") and the right side with respect to x (that's why we use "dx") :
\displaystyle\int{\left.{d}{y}\right.}=\int{\left({x}^{2}-{3}\right)}{\left.{d}{x}\right.}∫dy=∫(x2−3)dx
Then the answer is the same as before, but this time we have arrived at it considering the dy part more carefully:
\displaystyle{y}=\frac{{x}^{3}}{{3}}-{3}{x}+{K}y=3x3−3x+K
On the left hand side, we have integrated \displaystyle\int{\left.{d}{y}\right.}=\int{1}{\left.{d}{y}\right.}∫dy=∫1dy to give us y.
Similar questions