Question

a function y=f(x) satisfies xf'(x)-2f(x)=x^4f^2(x) ​

Answer 1

Step-by-step explanation:

tan30=

CD

AC

=>

CD

AC

=

3

1

=>LetAC=x,CD=x

3

AE=AC+CE=x+25

sin45=

2

1

=

AB

AE

=>

75

x+25

=

2

1

=> x=

2

75

2

−50

=27.85m

=> AC=27.85m,CD=48.18m

AD=

AC

2

+CD

2

=55.67m = lenght of string of second boy

solution.......

Answer 2

Answer:

f'(x)=24x^{-5}

Step-by-step explanation:

The formula for the derivative of $x$ given as $\frac{dy}{dx}$ or $(x)'$ $=1$

The derivative of $\frac{1}{x} = -\frac{1}{x^{2} }$

Let $u,v$ are two functions then the formula of integration is

$\int\limits {uv} \, dx =u\int\limits {v} \, dx -\int\limits {u'} \, ( \int\limits{v} \, dx )dx$

Given that,

$x\frac{dy}{dx} -2y=x^{4} y^{2}$

Dividing by $xy^{2}$,

we get

$\frac{1}{y^{2} } \frac{dy}{dx} -\frac{2}{xy} =x^{3}$----------------(1)

This is a Bernoulli differential equation

put

$-\frac{2}{y} =t\\\frac{2dy}{y^{2}dx } =\frac{dt}{dx}$

Equation (1) becomes

$\frac{1dt}{2dx} +\frac{t}{x} =x^{3} \\\frac{dt}{dx} +\frac{2t}{x} =2x^{3}$

I.F. = $e^{\int\limits \frac{2}{x}dx = e^{2lnx} \\ =e^{lnx^{2} } \\ =x^{2}$

so, general solution is given by

$t.x^{2} =\int\limits( {x^{2} .2x^{3} }) \, dx =2\int\limits {x^{5} } \, dx$

$x^{2} t=\frac{2x^{6} }{6} +c\\-\frac{2x^{2} }{y} =\frac{x^{6} }{3} +c\\-\frac{2}{y} =\frac{x^{4} }{3} +\frac{c}{x^{2} }$

If $x=1, y=-6$

then $c=0$

∴ The particular solution is

$-\frac{2}{y} =\frac{x^{4} }{3} \\y=-\frac{6}{x^{4} }$

That is $f(x)=-\frac{6}{x^{4} }$

then $f'(x)=24x^{-5}$