Answer:
20
Step-by-step explanation:
1. You "cross-multiply" 120 to 5 (120*5).
2. You divide 600 (120*5=600) to 30, which is (600/30).
3. You get the answer which is 20.
Answer:
ngl, he does have a decent amount of subs for a beginner
Step-by-step explanation:
I would if I was allowed to
Answer:
Verified
![y(x) = \frac{3Ln(x) + 3}{x}](https://tex.z-dn.net/?f=y%28x%29%20%3D%20%5Cfrac%7B3Ln%28x%29%20%2B%203%7D%7Bx%7D)
![y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}](https://tex.z-dn.net/?f=y%28x%29%20%3D%20%5Cfrac%7B3Ln%28x%29%20%2B%203%20-%203Ln%283%29%7D%7Bx%7D)
Step-by-step explanation:
Question:-
- We are given the following non-homogeneous ODE as follows:
![x^2y' +xy = 3](https://tex.z-dn.net/?f=x%5E2y%27%20%2Bxy%20%3D%203)
- A general solution to the above ODE is also given as:
![y = \frac{3Ln(x) + C }{x}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B3Ln%28x%29%20%2B%20C%20%20%7D%7Bx%7D)
- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.
Solution:-
- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.
![y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}](https://tex.z-dn.net/?f=y%27%20%3D%20%5Cfrac%7B%5Cfrac%7Bd%7D%7Bdx%7D%28%203Ln%28x%29%20%2B%20C%20%29%20.%20x%20-%20%28%203Ln%28x%29%20%2B%20C%20%29%20.%20%5Cfrac%7Bd%7D%7Bdx%7D%20%28x%29%20%20%7D%7Bx%5E2%7D%20%5C%5C%5C%5Cy%27%20%3D%20%5Cfrac%7B%5Cfrac%7B3%7D%7Bx%7D.x%20-%20%28%203Ln%28x%29%20%2B%20C%20%29.%281%29%7D%7Bx%5E2%7D%20%5C%5C%5C%5Cy%27%20%3D%20-%20%5Cfrac%7B3Ln%28x%29%20%2B%20C%20-%203%7D%7Bx%5E2%7D)
- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:
![-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3](https://tex.z-dn.net/?f=-%5Cfrac%7B3Ln%28x%29%20%2B%20C%20-%203%7D%7Bx%5E2%7D.x%5E2%20%2B%20%5Cfrac%7B3Ln%28x%29%20%2B%20C%7D%7Bx%7D.x%20%3D%203%5C%5C%5C%5C-3Ln%28x%29%20-%20C%20%2B%203%20%2B%203Ln%28x%29%20%2B%20C%3D%203%5C%5C%5C%5C3%20%3D%203)
- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.
- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:
![y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3](https://tex.z-dn.net/?f=y%28%201%20%29%20%3D%20%5Cfrac%7B3Ln%281%29%20%2B%20C%20%7D%7B1%7D%20%3D%203%5C%5C%5C%5C0%20%2B%20C%20%3D%203%2C%20C%20%3D%203)
- Therefore, the complete solution to the given ODE can be expressed as:
![y ( x ) = \frac{3Ln(x) + 3 }{x}](https://tex.z-dn.net/?f=y%20%28%20x%20%29%20%3D%20%5Cfrac%7B3Ln%28x%29%20%2B%203%20%7D%7Bx%7D)
- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:
![y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)](https://tex.z-dn.net/?f=y%283%29%20%3D%20%5Cfrac%7B3Ln%283%29%20%2B%20C%7D%7B3%7D%20%3D%201%5C%5C%5C%5Cy%283%29%20%3D%203Ln%283%29%20%2B%20C%20%3D%203%5C%5C%5C%5CC%20%3D%203%20-%203Ln%283%29)
- Therefore, the complete solution to the given ODE can be expressed as:
![y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}](https://tex.z-dn.net/?f=y%28x%29%20%3D%20%5Cfrac%7B3Ln%28x%29%20%2B%203%20-%203Ln%283%29%7D%7By%7D)
Answer:
under
Step-by-step explanation:
change the underminator, we have
7×(x-2)-4×(3x+5)=-3×7×4
7x-14-12x-20=-84
-5x=-50
x=10
Answer:
first setup your problem. it should look like this :