Question

Find the specific solution of the differential equation

Answer 1

@CausticSyndicalist @amistre64 @dan815 @rational @dtan5457 @freckles @HelpBlahBlahBlah @Michele_Laino @inowalst @sleepyjess @TheSmartOne @SithsAndGiggles @TuringTest @triciaal

Answer 2

we can try to separate the variables

Answer 3

namely:
\[\large \frac{{dy}}{{4y}} = \frac{{dx}}{{{x^2}}}\]

Answer 4

\[\frac{ 1 }{ x^2 } dx = 4y dy\]

Answer 5

@Michele_Laino

Answer 6

then we take the integral right?

Answer 7

no, please it is:
\[\large \frac{{dy}}{{4y}} = \frac{{dx}}{{{x^2}}}\]

Answer 8

then we have to take the integral of both sides

Answer 9

you should get this:
\[\large \frac{1}{4}\ln y = - \frac{1}{x} + C\]

Answer 10

yep got it! :)

Answer 11

ok!

Answer 12

next step is:
\[\large \ln y = - \frac{4}{x} + C\]

Answer 13

then multiply by e?

Answer 14

more precisely:
\[\large \ln y = - \frac{4}{x} + {C_1}\]
where C_1 is different from C

Answer 15

now, we have to apply the definition of logarithm, so we get:
\[\large y\left( x \right) = {C_1}\exp \left( { - \frac{4}{x}} \right)\]

Answer 16

exp?

Answer 17

yes!

Answer 18

what's that?

Answer 19

exp what is that, it was a question

Answer 20

more precisely:
\[\large y\left( x \right) = {C_2}{e^{ - 4/x}}\]
wher C_2 is different from C_1

Answer 21

ok, then what

Answer 22

What class is this for?

Answer 23

we have to apply the initial condition, namely y=e when x=-4, so we get:
\[\large e = {C_2}{e^{ - 4/ - 4}} = {C_2}e\]
what is C_2?

Answer 24

calculus

Answer 25

it is an Ordinary Differential Equation, namely ODE @FEARLESS_JOCEY

Answer 26

\[\large e = {C_2}e\]
what is C_2?

Answer 27

1?

Answer 28

ok!

Answer 29

finally, we have to replace thatvalue of C_2 into our solution, so we get:
\[\large y\left( x \right) = {C_2}{e^{ - 4/x}} = ...?\]

Answer 30

we have to substitute C_2=1

Answer 31

what do you get?

Answer 32

so C?

Answer 33

yes!

Answer 34

can u help me w/ another problem?

Answer 35

ok!

Answer 36

we have to solve your differential equation first

Answer 37

actually i posted the wrond problem

Answer 38

here, as before, we have to separate the variables, so we get:
\[\large \frac{{dy}}{y} = 2dt\]

Answer 39

yep

Answer 40

then, we have to take the integral of both sides, what do you get?

Answer 41

\[\frac{ y^2 }{ dy } = t + C\]

Answer 42

on the first, it should be noted that C is the only options whose derivative is

4(e^(-4/x))/x^2

and since y = e^(-4/x)

4y/x^2; so its either this or none

but; y(-4) = e^(-4/-4) = e, so its valid

this is assuming you know how to take a derivative to start with of course.

Answer 43

I got a different result @familyguymath

Answer 44

hint:
\[\large \frac{{dy}}{y} = \ln y\]

Answer 45

I can look at your answer as soon as possible @scarlettthecamel

Answer 46

so after a substitution we get:
\[\large \ln y = 2t + C\]

Answer 47

mult by e?

Answer 48

no, we have to apply the definition of logarithm, so we can write:
\[\large y\left( t \right) = {C_1}{e^{2t}}\]
where C_1 is another arbitrary constant which is different from C

Answer 49

more explanation:
we can write these steps:
\[\large y\left( t \right) = {e^{2t + C}} = {e^C}{e^{2t}} = {C_1}{e^{2t}}\]

Answer 50

please, tell me when I may continue

Answer 51

Okay, got it now

Answer 52

now, we have to apply the initial condition, namely y=60 when t=0,
so we can write:
\[\large 60 = {C_1}{e^0}\]
what is C_1?

Answer 53

hint:
\[\large {e^0} = 1\]

Answer 54

60

Answer 55

ok! So, our solution is:
\[\large y\left( t \right) = {C_1}{e^{2t}} = ...?\]

Answer 56

we have to replace C_1 with 60

Answer 57

A then?

Answer 58

that's right!

Answer 59

Can you help me with one more possibly?

Answer 60

ok!

Answer 61

The slope of the tangent line to a curve at any point (x, y) on the curve is x/ y. What is the equation of the curve if (3, 1) is a point on the curve?

x^2 + y^2 = 8
x + y = 8
x^2 - y^2 = 8
xy = 8

Answer 62

here we have to solve this differential equation:
\[\large \frac{{dy}}{{dx}} = \frac{x}{y}\]

Answer 63

now, we have to separate the variables, so we get:
\[\large y\;dy = x\;dx\]

Answer 64

is it ok?

Answer 65

Yes!

Answer 66

ok! Now we have to take the integral of both sides, so what do you get?

Answer 67

hint:
\[\large \int {z\;dz} = \frac{{{z^2}}}{2}\]

Answer 68

where z is any variable

Answer 69

hint:
\[\large \int {y\;dy} = \int {x\;dx} + C\]
please continue

Answer 70

is it

Answer 71

(/xz^2)/z + C