General solution for differentiation

Please can anyone help with these two equations.

I am asked to provide the general solutions.

a) 3^2 y^2dy/dx = 2 − 1

b) 2dy/dx + 3 = −2 − 5

Any help will be greatly appreciated!

 

a) To solve 3^2 y^2dy/dx = 2 − 1, we can first simplify it by dividing both sides by 3^2 y^2, which gives us:

dy/dx = (2 - 1)/(3^2 y^2)

dy/dx = 1/(9y^2)

Now we can separate the variables by multiplying both sides by 9y^2 and dx:

9y^2 dy = dx

Next, we integrate both sides:

∫9y^2 dy = ∫dx

3y^3 = x + C

where C is the constant of integration.

Therefore, the general solution to the differential equation is y = (x/3y^2) + C^(1/3).

b) To solve 2dy/dx + 3 = −2 − 5, we can first simplify it by subtracting 3 from both sides and dividing by 2:

dy/dx = -4/2

dy/dx = -2

Now we can integrate both sides with respect to x:

∫dy = ∫(-2)dx

y = -2x + C

where C is the constant of integration.

Therefore, the general solution to the differential equation is y = -2x + C.

 

a) 3^2 y^2dy/dx = 2 − 1
It's strange that you would divide by 3^2 y^2= 9 y^2 and then, later, multiply by it!
9y^2 dy/dx= 1
9y^2dy= dx
3y^3= x+ C
I would NOT write it as y= a function of both x and y. Leave it as 3y^3= x+ C or
write 3y^3- x= C.

b) 2dy/dx + 3 = −2 − 5
2 dy/dx+ 3= -7
2 dy/dx= -10 (NOT -4!)
dy= -5 dx
y= -5x+ C.

 

a)

Starting with 3^2y^2dy/dx=2−1,

We can simplify it to 9y^2dy/dx=1.

Now, we can separate variables:

9y^2dy=dx

Integrating both sides:

∫9y^2dy=∫dx

9/3y^3=x+C1

3y^3=x+C1

Solving for y:

y=(x+C1/3)^1/3

So, the general solution for equation (a) is:

y=(x+C1/3)^1/3

b)

For the second equation 2dy/dx+3=−2−5,

let's isolate dy/dx :

2dy/dx=−2−5−3

2dy/dx=−10

Now, divide by 2:

dy/dx =−5

This is a first-order linear differential equation, so its solution is straightforward. Integrating both sides:

∫dy=∫−5dx

y=−5x+C2

So, the general solution for equation (b) is:

y=−5x+C2

Hope this helps!



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