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Differentiation and Integration?
- Thread starter TJS1996
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I have moved this topic to the Calculus sub-forum, as it involves differentiation and integration.
Let's begin with the first problem. I am assuming you know the following rules:
a) \(\displaystyle \frac{d}{dx}(\sin(u(x)))=\cos(u)\frac{du}{dx}\)
b) \(\displaystyle \frac{d}{dx}(\cos(u(x)))=-\sin(u)\frac{du}{dx}\)
c) \(\displaystyle \frac{d}{dx}(kx^n)=knx^{n-1}\)
Can you apply these appropriately to each term?
Let's begin with the first problem. I am assuming you know the following rules:
a) \(\displaystyle \frac{d}{dx}(\sin(u(x)))=\cos(u)\frac{du}{dx}\)
b) \(\displaystyle \frac{d}{dx}(\cos(u(x)))=-\sin(u)\frac{du}{dx}\)
c) \(\displaystyle \frac{d}{dx}(kx^n)=knx^{n-1}\)
Can you apply these appropriately to each term?
chisigma
Well-known member
- Feb 13, 2012
- 1,704
If You have a differential equation written in the form...
$$f(y)\ dy = g(x)\ dx\ (1)$$
... F(*) is any primitive of f(*) and G(*) any primitive of g(*), then the solution is in the form...
$$F(y) = G(x) + c\ (2)$$
... where c is an arbitrary constant. Can Your equation be written in the form (1) and how?...
Kind regards
$\chi$ $\sigma$
- Jan 30, 2012
- 2,513
If $F'=f$, then $F$ is usually called an antiderivative, or indefinite integral, of $f$ in English. However, Wikipedia gives "primitive integral" as another possible synonym.
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Several months has gone by, and so I will now provide the solutions:
a) Differentiate the following with respect to $x$:
\(\displaystyle y=3\sin(4x)-5+5\cos\left(\frac{x}{3} \right)+4x\)
Differentiating term by term, and applying the chain rule as necessary, we find:
\(\displaystyle \frac{dy}{dx}=3\cos(4x)\cdot4-0-5\sin\left(\frac{x}{3} \right)\cdot\frac{1}{3}+4\)
\(\displaystyle \frac{dy}{dx}=12\cos(4x)-\frac{5}{3}\sin\left(\frac{x}{3} \right)+4\)
b) Solve the following IVP:
\(\displaystyle \frac{dy}{dx}=4\cos(2x)+6\) where \(\displaystyle y(0)=7\)
Switching dummy variables of integration, and using the initial values as the limits of integration, we may write:
\(\displaystyle \int_{y(0)}^{y(x)}\,du=2\int_0^x2\cos(2v)+3\,dv\)
\(\displaystyle y(x)-y(0)=2\left[\sin(2v)+3v \right]_0^x=2\left(\sin(2x)+3x \right)\)
And so the solution satisfying the given conditions is:
\(\displaystyle y(x)=2\left(\sin(2x)+3x \right)+7\)
There it is...