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olga nikolaevna [1]
4 years ago
7

X ^ (2) y '' - 7xy '+ 16y = 0, y1 = x ^ 4

Mathematics
1 answer:
nignag [31]4 years ago
4 0
Standard reduction of order procedure: suppose there is a second solution of the form y_2(x)=v(x)y_1(x), which has derivatives

y_2=vx^4
{y_2}'=v'x^4+4vx^3
{y_2}''=v''x^4+8v'x^3+12vx^2

Substitute these terms into the ODE:

x^2(v''x^4+8v'x^3+12vx^2)-7x(v'x^4+4vx^3)+16vx^4=0
v''x^6+8v'x^5+12vx^4-7v'x^5-28vx^4+16vx^4=0
v''x^6+v'x^5=0

and replacing v'=w, we have an ODE linear in w:

w'x^6+wx^5=0

Divide both sides by x^5, giving

w'x+w=0

and noting that the left hand side is a derivative of a product, namely

\dfrac{\mathrm d}{\mathrm dx}[wx]=0

we can then integrate both sides to obtain

wx=C_1
w=\dfrac{C_1}x

Solve for v:

v'=\dfrac{C_1}x
v=C_1\ln|x|+C_2

Now

y=C_1x^4\ln|x|+C_2x^4

where the second term is already accounted for by y_1, which means y_2=x^4\ln x, and the above is the general solution for the ODE.
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Hey there!

Same question is answered by me correctly.

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<u>Note:</u>

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