How do I solve the simultaneous equations x^2+y^2 =5 and y=3x+1?
x^2 + y^2 = 5 -(1)y = 3x+ 1 -(2)Put the value of y from eq (2) in eq (1)-x^2+ (3x+1)^2=5x^2+ 9x^2+ 1+6x = 510x^2 + 6x -4 = 010x^2 + (10 - 4)x - 4 = 010x^2 + 10 x - 4 x - 4 = 010x (x+ 1) -4 (x+1) = 0(10x-4)(x+1) =0Hence x= (-1), (2/5)For x= -1y= 3×(-1)+1y= -2For x=2/5y=3×(2/5)+1y=11/5
Solve simultaneous equation: y - 3x + 2=0, y^2 - x - 6x^2?
y - 3x + 2=0 => y = 3x - 2 → sub. y in the equation below y^2 - x - 6x^2 = 0 → assuming the equation = 0 (3x - 2)^2 - x - 6x^2 = 0 9x^2 - 12x + 4 - x - 6x^2 = 0 3x^2 - 13x + 4 = 0 (3x - 1)(x - 4) = 0 x = 1/3 , 4 → sub. y = 3x - 2 y = -1 , 10 hence: the solutions (points of intersection) are: (x , y) = (1/3 , -1) & (4 , 10)
How can I solve this system of equations? y= 2x^2 -3 and y= 3x -1
Equating them next to each other (because invariably, y = y for all real numbers y and I assume this to be a real-valued function),2x² - 3 = 3x - 12x² - 3x - 2 = 0(2x + 1)(x - 2) = 0For this product to be 0, either factor needs to be equal to 0. This is true because they are reduced to polynomials of degree 1:2x + 1 = 0 or x - 2 = 0→ x = -½ or x = 2Kind regards,Zane Heyl
How do I plot a the curve with the equation of y=3x^2/3-2x?
All these answers that involve using technology. Here's a simpler way to do it using graph paper, a pencil, and your head:Solve for the roots. This involves setting y to 0 and solving for the x values. Plug in values for x and get the corresponding values for y(x). I would suggest plugging in integers between -10 and 10. Now connect the dots until you get a graph. This is your plot. Remember that you need to enter enough data points depending on the highest degree of the polynomial to ensure you have the right shape.
How do I find "dy/dx" with the equation "y = 3x^4 + 2x^2 + 10"?
Let f(x) = 3x^4 + 2x^2 + 10f(x+h) = 3(x^4 +4x^3 h + 6x^2 h^2 + 4xh^3 + h^4) +2(x^2 + 2xh + h^2) + 10f(x+h) - f(x) = 12x^3 h + + 18x^2 h^2 + 12xh^3 + 3h^4) + 4xh +2h^2(f(x+h)-f(x))/h = 12x^3 +18x^2 h + 12xh^2 +3h^3 +4x +2hTake the limit of this last expression as h goes to 0. You getf’(x) = 12x^3 + 4x
Y=x+2 y=3x-2 solution?
You're being told that y is equal to both x+2 and 3x-2. So you can set these two expressions equal to each other. x+2 = 3x-2 Solve this equation for x, then use that value of x in either of the original equations to find the value of y.
What is the solution to this ODE: (3x^2 + 4xy + y^2) dx + (2x^2 + 2xy + 9) dy = 0?
[math](3x^2 + 4xy + y^2) dx + (2x^2 + 2xy + 9) dy = 0[/math]Equation of the form [math]\quad M(x,y) dx + N(x,y) dy = 0[/math] is exact when [math]\quad\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}[/math]In that case, solution is of the form [math]F(x,y) = C[/math], where[math]\quad M(x,y) = \frac{\partial F}{\partial x}[/math] [math]\quad N(x,y) = \frac{\partial F}{\partial y}[/math][math]M(x,y) = 3x^2 + 4xy + y^2 \Longrightarrow \frac{\partial M}{\partial y} = 4x + 2y[/math] [math]N(x,y) = 2x^2 + 2xy + 9 \Longrightarrow \frac{\partial N}{\partial x} = 4x + 2y[/math]Equation is exact, so we find [math]F(x,y)[/math] by integrating:[math]F(x,y) = \int M(x,y) dx = \int (3x^2 + 4xy + y^2) dx = x^3 + 2x^2y + xy^2 + g(y)[/math] [math]F(x,y) = \int N(x,y) dy = \int (2x^2 + 2xy + 9) dy = 2x^2y + xy^2 + 9y + h(x)[/math]Comparing both results, we see that [math]g(y) = 9y, h(x) = x^3,[/math] and therfore:[math]F(x) = x^3 + 2x^2y + xy^2 + 9y[/math]Solution:[math]\boxed{\boldsymbol{x^3 + 2x^2y + xy^2 + 9y = C}}[/math]
How do I solve y'''''' = 3x^2 - sin2x?
Integrate 6 times, which is not difficult for this particular set.[math]y’’’’’’=3x^2-\sin 2x[/math][math]u=2x;du=2dx;dx=du/2[/math][math]y’’’’’=\displaystyle\int 3x^2\;dx-\displaystyle \int\frac 12\sin u\;du[/math][math]=x^3+\frac 12\cos 2x+C_1[/math]From here we can seeOur polynomial [math]x^3[/math] is just going to get more and more factors in its divisor.Our trig term is just going to be changing function and getting divided by 2.We have a constant of integration, and five others, that also need integrating.So, to simplify the pattern we are going to express our function in form that makes the pattern easier to handle:[math]y’’’’’=\dfrac {3!x^3}{3!}+\dfrac {\cos 2x}{2^1}+C_1[/math]From there, it's pretty simple to handle:[math]y’’’’=\dfrac {3!x^4}{4!}+\dfrac {C_1x}{1!}+\dfrac {\sin 2x}{2^2}+C_2[/math][math]y’’’=\dfrac {3!x^5}{5!}+\dfrac {C_1x^2}{2!}+\dfrac {C_2x}{1!}-\dfrac {\cos 2x}{2^3}+C_3[/math][math]y’’=\dfrac {3!x^6}{6!}+\dfrac {C_1x^3}{3!}+\dfrac {C_2x^2}{2!}+\dfrac{C_3x}{3!}-\dfrac {\sin 2x}{2^4}+C_4[/math][math]y’=\dfrac {3!x^7}{7!}+\dfrac {C_1x^4}{4!}+\dfrac {C_2x^3}{3!}+\dfrac{C_4x^2}{2!}+\dfrac {C_4x}{1! }+\dfrac {\cos 2x}{2^5}+C_5[/math]And finally:[math]y=\dfrac {3!x^8}{8!}+\dfrac {C_1x^5}{5!}+\dfrac {C_2x^4}{4!}+\dfrac{C_4x^3}{3!}+\dfrac {C_4x^2}{2!}+C_5x+\dfrac {\sin 2x}{2^6}+C_6[/math]