X = (2y+1)/(y+2) solve for y?
Y=y+2, then Y-2=y. Now x=(2Y-3)/Y=2-3/Y, or x-2=-3/Y which is same as 3/(2-x)=Y=y+2. And so, y=(2x-1)/(2-x).
How do I solve this: x^2 (dy/dx) = y^2 + 2y + 1 , where x>0?
This is a separable differential equation.The right side of the equation can be simplified further by completing the square.Thus[math]\displaystyle x^{2}\frac{dy}{dx}=(y+1)^{2}[/math][math]\displaystyle \frac{1}{(y+1)^{2}}dy=\frac{1}{x^{2}}dx[/math]Let’s integrate[math]\displaystyle \int \frac{1}{(y+1)^{2}}dy=\int \frac{1}{x^{2}}dx[/math][math]\displaystyle -(y+1)^{-1}=-x^{-1}+C[/math][math]\displaystyle y+1=\frac{1}{x^{-1}-C}[/math][math]\displaystyle y=\frac{1}{x^{-1}-C}-1[/math]
How to solve x2-2y=1?
"Solve, solve, solve"! That's what everybody says. What do you mean by "solve"? There is nothing to "solve" here. One equation, two variables. You cannot "solve" such. You can do nothing with it.
How can I solve inequality of √x^2+√y^2 =5 and 3x-2y=1?
Case 1 if both x and y > 0 than equation isX+y=53x-2y=1 solve this we getX=11/5Y=14/5Case 2 if x and y both< 0 than equation is-x-y = 4 and 3x-2y=1Now you can solve this.Case 3 if. x>0. And y<0 than equation isX-y=5 and. 3x-2y=1Now you can solve thisCase 4If. X<0 and y > 0 than equation is-x + y =5 and. 3x-2y=1Now solve by your self
Solve for y 1/5(2y-1_-2=1/2(3y-5)+3 I got y= -2 1/11 is this correct?
i hope your equation is like this: 1/5(2y-1)-2 = 1/2(3y-5) +3 First multiply by 10: 2(2y-1)-20=5(3y-5)+30 4y-2-20=15y-25+30 4y-15y=-25+30+2+20 -11y=27 y=-27/11=-2 5/11
Solve the simultaneous equation: x - 2y = 1 and x^2 + y^2 + 2y = 33?
I would use substitution to solve the two equations. In that case you would do the following: Rearrange the first equation to solve for x: x = 1 + 2y Now plug that back into the second equation: (1 + 2y)^2 + y^2 + 2y = 33 Expand the (1 + 2y)^2: (1 + 2y)(1 + 2y) 1 + 2y + 2y + 4y^2 Put it all together: (1 + 4y + 4y^2) + y^2 + 2y = 33 Thus: 5y^2 + 6y + 1 = 33 And now solve for y: 5y^2 + 6y - 32 = 0 (5y+16)(y-2) = 0 y = 2 or -3.2 Now you just plug in the y values into the first equation to get the possible x values: x - 2(2 or -3.2) = 1 x = 5 or -5.4 y = 2 or -3.2 That should be it.
How do I solve the system of equation x-2y=-1 , 2x^2+x-1=y^2-2y?
This can be solved using simultaneous equations.x - 2y = -1, set this as your first equation and set 2x^2 + x-1 = y^2 -2y as your second.From this you can get x = 2y - 1 by moving the 2y to the RHS.Substitute this into the second equation and you get.2(2y-1)^2 + (2y-1) - 1 = y^2 - 2y2(4y^2 -4y +1) + 2y - 2 = y^2 - 2y8y^2 - 8 y + 2 + 2y - 2 = y^2 - 2y7y^2 - 4y = 0y (7y - 4) = 0This means that y can be either 0 or 4/7.You substitute these y values back into the first equation of x = 2y - 1 and get the respective x values.x = -1 , y = 0x = 1/7 , y = 4/7
What is the particular solution for the equation y’’+3y’+2y= 1/ (1+e^x)?
Q: What is the particular solution for the equation y’’+3y’+2y= 1/ (1+e^x)?[math]\textrm{Use the Characteristic Equation:} \tag*{} \\[/math][math]r^2+3r+2 = 0 \tag*{} \\[/math][math](r+1)(r+2) = 0 \tag*{} \\[/math][math]r = -2, -1 \tag*{} \\[/math][math]\textrm{The General Solution is:} \tag*{} \\[/math][math]y = c_1e^{-x}+c_2e^{-2x} \tag*{} \\[/math][math]\textrm{Find the Full Solution using Variation by Parameters:} \tag*{} \\[/math][math]y_1 = e^{-x} \tag*{} \\[/math][math]y_2 = e^{-2x} \tag*{} \\[/math][math]\mathcal{W} = \begin{vmatrix} e^{-x} & e^{-2x} \\ -e^{-x} & -2e^{-2x} \end{vmatrix} = -2e^{-3x}+e^{-3x} = -e^{-3x} \tag*{} \\[/math][math]v_1(x) = -\displaystyle \int \dfrac{f(x)y_2}{\mathcal{W}} \,\mathrm dx \tag*{} \\[/math][math]v_1(x) = \displaystyle \int \dfrac{e^x}{1+e^x} \,\mathrm dx \tag*{} \\[/math][math]v_1(x) = \ln \left(1+e^x \right)+C \tag*{} \\[/math][math]v_2(x) = \displaystyle \int \dfrac{f(x)y_1}{\mathcal{W}} \,\mathrm dx \tag*{} \\[/math][math]v_2(x) = \displaystyle \int \dfrac{-e^{2x}}{1+e^x} \,\mathrm dx \tag*{} \\[/math][math]v_2(x) = \ln \left(1+e^x \right)-e^x+C \tag*{} \\[/math][math]\textrm{The Particular Solution will be in the form of:} \tag*{} \\[/math][math]y_p = v_1(x)e^{-x}+v_2(x)e^{-2x} \tag*{} \\[/math][math]y_p = e^{-x} \ln \left(1+e^x \right)+e^{-2x} \ln \left(1+e^x \right)-e^{-x} \tag*{} \\[/math][math]\therefore y_p = e^{-x} \ln \left(1+e^x \right)+e^{-2x} \ln \left(1+e^x \right) \tag*{} \\[/math]
How do I factorise x2-y2-2y-1?
Like this.x²-y²-2y-1=x²-(y²+2y+1)=x²-(y+1)²=(x+y+1)(x-y-1)