General solution for differentiation
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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.
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.
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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.
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.
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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!
In working through this mathematical problem, it's clear that precision and accuracy are key. For those seeking assistance or guidance with such calculations, I will suggest you to try MathsAssignmentHelp.com that can be incredibly valuable. Also, you can contact them at +1 (315) 557-6473.
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!
In working through this mathematical problem, it's clear that precision and accuracy are key. For those seeking assistance or guidance with such calculations, I will suggest you to try MathsAssignmentHelp.com that can be incredibly valuable. Also, you can contact them at +1 (315) 557-6473.
