Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If $a{y^4} = {(x + b)^5}$ then show that $5y{y''} = {(y')^2}$

seo-qna
Last updated date: 16th Jul 2024
Total views: 347.7k
Views today: 3.47k
Answer
VerifiedVerified
347.7k+ views
Hint: First we have to define the terms we need to solve in the problem. Since this problem is based on the method of the differentiation which is if we differentiate ${x^n}$ then we get $n{x^{n - 1}}$ so using this concept we will solve it further. Also, differentiation is with respect to the variables or values or numbers. This may also be called the inverse of the integration.

Complete step-by-step solution:
From the given question we have $a{y^4} = {(x + b)^5}$ first place the value $a$into the right hand side; then we get ${y^4} = \dfrac{{{{(x + b)}^5}}}{a}$ (a and b are constants here so after acting the differentiation does not affects the constant ) In next step turn the square values into square root to the right hand side or take common square root on both side we get; $y = {(\dfrac{1}{a})^{\dfrac{1}{4}}}{(x + b)^{\dfrac{5}{4}}}$ now we may assume that constant term into C
And hence $y = c{(x + b)^{\dfrac{5}{4}}}$ and since (${(\dfrac{1}{a})^{\dfrac{1}{4}}}$=C)
From the given definition we can further solve different ion of the value with respect to x;
First differentiation with respect to x we have; $y' = c \times \dfrac{5}{4}{(x + b)^{\dfrac{1}{4}}}$(the power will come down also the new power minus one) now again differentiation with respect to x we have;
${y''} = \dfrac{5}{4}c \times \dfrac{1}{4}{(x + b)^{\dfrac{{ - 3}}{4}}} $
$= \dfrac{{5c}}{{16}}{(x + b)^{\dfrac{{ - 3}}{4}}}$ (the power will come down also the new power minus one)
Hence, we have first and second differentiation of the given problem
Now from the problem take the left-hand side which is $5y{y''}$ and substitute the first and second differentiation values we get
$5y{y''} = 5c{(x + b)^{\tfrac{5}{4}}} \times \dfrac{{5c}}{{16}}{(x + b)^{\dfrac{{ - 3}}{4}}}$ and further solving by cancelling the terms we get;
$5yy{}'' = \dfrac{{25{c^2}}}{{16}}{(x + b)^{\dfrac{1}{2}}}$
Similarly, from the right side given equation ${(y')^2}$ and substitute the first differentiation we get;
${(y')^2} = {[c \times \dfrac{5}{4}{(x + b)^{\dfrac{1}{4}}}]^2} = \dfrac{{25{c^2}}}{{16}}{(x + b)^{\dfrac{1}{2}}}$
Thus clear the right and left side equations given the same value and hence $5y{y''} = {(y')^2}$.

Note: Here $x,y$are the variables so differentiation will only act on them and $a,b$are the constants so it won’t affect these values.
Also, differentiation of ${x^2} = 2{x^{2 - 1}} = 2x$ and ${x^{ - 2}} = - 2{x^{ - 3}}$;
Integration is the inverse function of differentiation. There is no other easy method to solve this problem.