Find explicit solution for ivp dy/dx = 4x^3y - y, y(1)= -3?
This a separable 1st order DE. So the steps are: 1. separate variables: dy/dx = 4x³ · y - y <=> dy/dx = (4x³ - 1) · y <=> (1/y) dy = (4x³ - 1) dx 2. integrate ∫ (1/y) dy = ∫ (4x³-1) dx => ln(y) + c1 = x^4 - x + c2 (you can merge the two constants to one constant c = c2 -c1) => ln(y) = x^4 - x + c => y = e^(x^4 - x + c) <=> y = C · e^(x^4 - x) (where C = e^c) 3. use initial condition to evaluate the constant y(1) = -3 <=> C·e^(1^4 - 1) = -3 <=> C = -3 Hence the solution of this initial value problem is y = -3·e^(x^4 - x)
How solve an explicit solution of the given initial value problem?
x^2 dy/dx = y - xy divide both sides by x^2 x^2 dy/dx = y(1-x) separate the variables dy/y = (1-x)/x^2 dx dy/y = (1/x^2 - 1/x) dx Integrate both sides ln y = -1/x - ln(x) + C1 y = e^[ -1/x - ln(x) + C1] y = e^(-1/x) e^ln(x^(-1)) e^C1 y = C e^(-1/x) /x , where C = e^C1 y = C e^(-1/x) / x y(-1) = -1 x=-1 ; y=-1 -1 = C e^(1) / (-1) -1 = -C e C = 1/e y = e^(-1/x) / xe y = e^(-1/x -1) / x
How do I solve the following problem and what type of equation is it? (3x^2-2y) dx+ (x^2y+2x) dy=0 y(-1) =4
How do I solve the following problem and what type of equation is it? (3x^2-2y) dx+ (x^2y+2x) dy=0 y(-1) =4A2AThis part is a differential equation:(3x^2-2y) dx+ (x^2y+2x) dy=0…and this part is an initial value:y(-1) =4The combination of the 2 parts is called an initial value problem (IVP).The solution to a differential equation is a function (not a number or variable).If you re-write that differential equation as:dy/dx = -(3x^2-2y) / (x^2y+2x)…then the function y=f(x) whose derivative is equal to -(3x^2-2y) / (x^2y+2x) is the “general” solution.Once you’ve found f(x), you can set f(-1) =4 and solve that to obtain the arbitrary constant in the general solution to obtain the “particular” solution… which is the solution of the IVP.Some (most) differential equations cannot be solved analytically. In that case, you can use numerical methods together with the provided initial value to solve for numerical values of y vs x..Here’s how to write the differential equation in WxMaxima:Here’s how to solve for the General Solution:Let’s take the derivative of GS to make sure it equals DE:Yup.Now solve for the Particular Solution:Solve the implicit Particular Solution explicitly for y:The above is the solution to the given Initial Value Problem.Let’s evaluate PSy at x=-1 just to make sure it’s equal to 4:Yup.Finally, here’s a plot of the IVP solution PSy:
Please provide the explicit solution for the IVP. dy/dx + 4y - e^-x = 0?
Rewrite in standard form: dy/dx + 4y = e^(-x). Multiply both sides by the integrating factor e^(∫ 4 dx) = e^(4x): e^(4x) dy/dx + 4e^(4x) y = e^(3x) ==> (d/dx) (e^(4x) y) = e^(3x) Integrate both sides: e^(4x) y = (1/3) e^(3x) + C ==> y = (1/3) e^(-x) + Ce^(-4x). To find C, use y(0) = 5/3: 5/3 = 1/3 + C ==> C = 4/3. So, y = (1/3) e^(-x) + (4/3) e^(-4x). ---------------- I hope this helps!
Find an explicit solution of the given initial-value, determine exact interval I : sinxdx+ydy=0 ; y(0)=1?
I'm not sure how you got the equation equal to 1, since it's given to be equal to 0. sinx dx + y dy = 0 sinx + y (dy/dx) = 0 y (dy/dx) = -sinx y dy = -sinx dx ∫ y dy = ∫ (-sinx) dx 1/2 y² = cosx + C y² = 2cosx + C Given that y(0) = 1, (1)² = 2 cos(0) + C 1 = 2 + C C = -1 The solution is y² = 2cosx - 1 (implicit form)
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]
What is the solution to [math]\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{1}{y^2}[/math]?
You say you “just want to know the solution and not the method.” As I begin to write, I see that two solution methods (essentially the same) have been presented (by Anirban Ghoshal and Awnon Bhowmik), neither of which has culminated in a complete solution (by complete, I mean up to two arbitrary constants of integration together with assumptions that must be made about the nature of their values, and of the possible values of the variable [math]y[/math]). So to give some closure here, I’ll try and complete their solution for you, as I believe their method to be correct. Since Anirban carried it the furthest, and also specified more completely the assumptions he made ([math]C > 0,\,\, y > 0[/math]), I will begin with his first integral of the ODE (the same as that found by Awnon, and where he left his solution):[math]\displaystyle\int\sqrt{\dfrac{y}{Cy-2}}\,dy = x + C_1. [/math]Following Anirban, make the change of variable to [math]y = t^2\,\,[/math], [math]dy = 2t\,dt[/math], bringing the integral to[math]\displaystyle\int \dfrac{2t^2}{\sqrt{Ct^2 - 2}}\,dt = x + C_1.[/math]The integral on the left can be integrated by parts with [math]u = (Ct^2 - 2)^{1/2}/C\,[/math], [math]\,\,dv = 2\,dt[/math], [math]\,du = t(Ct^2 - 2)^{-1/2}\,[/math], [math]\,v = 2t[/math], to obtain[math]\displaystyle\dfrac{2t}{C}\sqrt{Ct^2 - 2} - \dfrac{2}{C}\int \sqrt{Ct^2 - 2}\,dt = x + C_1. [/math]The final integral involving the square root is messy, but a change of variables to [math]z = \sqrt{C/2}\,t\,\,[/math], [math]dz = \sqrt{C/2}\,dt\,\,[/math] brings it to[math]\displaystyle\dfrac{2t}{C}\sqrt{Ct^2 - 2} - \dfrac{4}{C \sqrt{C}} \int \sqrt{z^2 - 1}\,dz = x + C_1.[/math]It just gets worse, because now, as far as I can tell, you need to introduce the secant function via [math]z = \sec{\theta}[/math], ending up with an integral of [math]\sec^3{\theta} - \sec{\theta}[/math]. Rather than go through all of that, I’ll just give you the final result, which is apparently what you really want to know:[math]\displaystyle\dfrac{t}{C}\sqrt{Ct^2 - 2} + \dfrac{2}{C\sqrt{C}}\ln{\big(Ct + \sqrt{C} \sqrt{Ct^2 - 2}\,\big)} = x + C_1,[/math]or, in terms of the original variable [math]y = t^2[/math]:[math]\displaystyle\dfrac{\sqrt{y}}{C}\sqrt{Cy - 2} + \dfrac{2}{C\sqrt{C}}\ln{\big(C\sqrt{y} + \sqrt{C} \sqrt{Cy - 2}\,\big)} = x + C_1.[/math]