With Work How Do You Solve 3xy- 2x 4xy

How do I solve (1+x^2) dy/dx + 3xy= 5x with y(1) =2?

[math]\begin{align}(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+3xy&=5x\\(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}&=5x-3xy\\(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}&=x(5-3y)\\\int\dfrac{\mathrm dy}{5-3y}&=\int\dfrac x{1+x^2}\mathrm dx\\-\dfrac13\ln|5-3y|&=\dfrac12\ln|1+x^2|+\ln C\\-2\ln|5-3y|&=3\ln|1+x^2|+6\ln C\\\ln|5-3y|&=-\dfrac32\ln|1+x^2|-3\ln C\\\ln|5-3y|&=-\ln|K(1+x^2)^{\frac32}|\qquad[\because K=3\ln C]\\\dfrac1{5-3y}&=K(1+x^2)^{\frac32}\\\hline\text{Using }y(1)=2\\\dfrac1{5-6}&=2\sqrt2K\\K&=-\dfrac1{2\sqrt2}\\\hline\dfrac1{5-3y}&=-\dfrac1{2\sqrt2}(1+x^2)^{\frac32}\\3y-5&=\dfrac{2\sqrt2}{(1+x^2)^{\frac32}}\\y&=\dfrac{\dfrac{2\sqrt2}{(1+x^2)^{\frac32}}+5}3\\y(x)&=\dfrac{2\sqrt2+5(1+x^2)^{\frac32}}{3(1+x^2)^{\frac32}}\end{align}\tag*{}[/math]

How do I solve for x and y in x + y =5 and xy=6?

Let’s first solve this problem by inspection. Notice that x and y can’t both be negative because then their addition will not produce the positive result required by the equation x + y = 5. And one of the unknowns cannot be negative (and the other positive) because then the product will not produce a positive result as required by the equation x y = 6. So x and y must both be positive and add up to 5. So the candidate solutions for the equation x + y = 5 are (0, 5), (1, 4) and (2, 3). Of these, only the pair (2, 3) will satisfy the equation x y = 6. So by inspection, we have that the solution is x = 2 and y = 3 as well as x = 3 and y = 2.Now let’s solve the problem analytically. From the first equation, we have that x = 5 – y. Plug this into the second equation, collect terms, and simplify to get y^2 – 5y + 6 = 0. This is a quadratic equation of the form ay^2 + by + c = 0, whose two solutions can be written as y = {-b ± (b^2 – 4ac)^1/2 }/2a. Substitute a =1, b = -5 and c = 6 into this result to find that our quadratic equation for y has two solutions: y = 3 and y = 2. Substitute these solutions for y back into the equation x = 5 – y to find that x = 2 and x = 3. This is the same result we found above by inspection.

Can I solve y''+ay'+by=0 using the power series?

That depends on what a and b are. If they are constants, they you're sure to be able to solve it by the power series method about any finite point.If a=a(x) and b=b(x) are variable coefficients, then you must look out for singularities of the equation. In that case, the generalized power series method works (atleast gives you one solution) if you're finding it about an ordinary point or a non-essential singularity.

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 would I solve [math]x^2 -5xy + 6y^2 = 0?[/math] Give steps please!

Looking at this question with 2 variables is very confusing.  Take y out (for now, don't question me.)Here's our new equation:x^2 - 5x + 6 = 0Now you can either use the quadratic equation, use the box method, or use any other method to solve for the roots and factor.The factored equation is:(x-2)(x-3) = 0From here we see that if we simply put a 'y' after the -2 and the -3, we get(x-2y)(x-3y) = 0which multiplies tox^2 - 5xy +6y^2 = 0which is the original equation, which means our roots are correct.Continuing from (x-2y)(x-3y) = 0we see that (x-2y) = 0 OR (x-3y) = 0 because 0 times anything equals 0.x-2y = 0x-3y = 0After setting each parentheses equal to 0, we can solve for x:x = 2yx = 3yThere are your answers.  If I made anything confusing or had typos, blame my unwillingness to get sleep. :/

How do I solve these simultaneous equations, x=3+2y, x²+2y²=27 and 2x²+xy+y²=22, x+y=1?

Let’s start with the first:x = 3+2yx^2 + 2y^2 = 27Firstly, we need to align both equations so that they use similar terms, we can create an equation that contains x^2 and y^2 by squaring both sides of x = 3+ 2y(x)^2 = (3+ 2y)^2x^2 = 4y^2 + 12y + 9Then, we can get both equations to parallel each other by altering x^2 + 2y^2 = 27 so that x^2 is alone:x^2 + 2y^2 = 27 therefore x^2 = -2y^2 +27No we can subtract one equation from the other:x^2 = 4y^2 + 12y + 9x^2 = -2y^2 + 27Therefore0 = 6y^2 + 12y - 18This equation can be simplified by dividing all the elements by 6, leaving:y^2 + 2y - 3 = 0This factorises to:(y + 3)(y - 1) = 0Which results in two y values: -3 and 1From these values, you can create two values of x by using the equation x = 3 + 2y:Let y = -3:x = 3 + 2(-3)x = 3 - 6 = -3This can be verified using the other original equation:x^2 + 2y^2 = 27(-3)^2 + 2(-3)^2 = 279 + 2(9) = 279 + 18 = 2727 = 27For the other value:Let y = 1:x = 3 + 2yx = 3 + 2(1)x = 3 + 2 = 5Verify:x^2 + 2y^2 = 27(5)^2 + 2(1)^2 = 2725 + 2 = 2727 = 27Considering how long it took to type this all out, I won’t do the second equation. Similar steps and processes are used though.