Differential Equation- HELP PLEASE!!!!!

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elijahbrooke

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Feb 1, 2012
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a) y (d²y/dx²) = 3(dy/dx)²

Also I'll put these ones up as well, it's homework but just in case I get stuck, it'll save me starting multiple threads...

b) (3y-2) (d²y/dx²) = 3 (dy/dx)²

c) (d²s/dt²)+0.1(ds/dt)², using the substitution (ds/dt) = v

d) x (d²y/dx²) + 2(dy/dx) = x ln(x)

With steps as well? Thanks.
 
elijahbrooke said:
a) y (d²y/dx²) = 3(dy/dx)²

y (d²y/dx²) = 3(dy/dx)²

(d²y/dx²) / (dy/dx) = 3(dy/dx)/y

Now integrate both sides and continue.....

Also I'll put these ones up as well, it's homework but just in case I get stuck, it'll save me starting multiple threads...

b) (3y-2) (d²y/dx²) = 3 (dy/dx)²

c) (d²s/dt²)+0.1(ds/dt)², using the substitution (ds/dt) = v

d) x (d²y/dx²) + 2(dy/dx) = x ln(x)

With steps as well? Thanks.
Click to expand...

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
Bit harsh but Subhotan is probably right. If you can't even START these problems then you are in big trouble. Not attending class (especially something like calculus) is just not a good idea. Although calculus breaks down into lots of sections, differential equations often build on top of each other so if you miss the first or second or third lecture you will find it hard to pick up, because you would of missed steps out in the harder questions.

Subhotosh has given you a hint for the first one, can you integrate both sides that he has given?
 
d can be done by using Variation of Parameters. First, divide through by x.

\(\displaystyle \displaystyle y''+\frac{2}{x}y'=ln(x)\)

Now, make the sub \(\displaystyle y=x^{m}\)

\(\displaystyle m(m-1)x^{m-2}+2mx^{m}=ln(x)\)

The auxiliary equation, \(\displaystyle m(m+1)=0\) has roots \(\displaystyle x=0, \;\ x=-1\)

\(\displaystyle y_{c}=C_{1}+C_{2}x^{-1}\)

Now, identify \(\displaystyle f(x)=ln(x), \;\ y_{1}=1, \;\ y_{2}=x^{-1}\) and perform the Wronskian.

If you have a DE text, look under Variation of Parameters.

Is c complete?. It is not an equation, but an expression.
 
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