Ex 94, 12 Find a particular solution satisfying the given condition 𝑥 𝑥2−1 𝑑𝑦𝑑𝑥=1;𝑦=0 When 𝑥=2 𝑥 𝑥2−1 dy = dx dy = 𝑑𝑥𝑥(𝑥2 − 1) Integrating both sides 𝑑𝑦 = 𝑑𝑥𝑥(𝑥2 − 1) 𝑦 = 𝑑𝑥𝑥(𝑥 1)(𝑥 − 1) We can write integrand as 1𝑥(𝑥 1 1x^2 dy/dx = 1y^2 Homework Equations The Attempt at a Solution if I clean this up a little, I would get 1/ (1x^2) dx = 1/ (1y^2) dy correct? Ex 53, 11 Find 𝑑𝑦/𝑑𝑥 in, 𝑦 = cos–1 ((1− 𝑥^2)/( 1 𝑥2 )) , 0 < x < 1 𝑦 = cos–1 ((1− 𝑥^2)/( 1 𝑥2 )) Putting x = tan θ y
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X^2(x^2-1)dy/dx x(x^2 1)y=x^2-1
X^2(x^2-1)dy/dx x(x^2 1)y=x^2-1-Why create a profile on Shaalaacom? But if I expand the bracket $(xy)^2$ before integrating I will get $$\varnothing_1=\int Mdx=\int (xy)^2dx=\int (x^22xyy^2)dx=\frac{x^3}{3}xy^2x^2y$$ Wich will lead to the solution $$\varnothing=\varnothing_1\varnothing_2=\frac{x^3}{3}xy^2x^2yy=Constant$$ What is the wrong step ?
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Simple and best practice solution for (x2)dx4(xy1)dy=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it The line that is normal to the curve x^2=2xy3y^2=0 at(1,1) intersects the curve at what other point? y= sqrt ( 2 sqrt(x^21) ) IF its actually x2ysqrt(x^21)dy/dx = 0 then we can say that 2ysqrt(x^21)dy/dx = x So 2ydy/dx = x/sqrt(x^21) Integrating both sides int \2 ydy/dx \ dx=int \ x/sqrt(x^21) \ dx 2 int \ d/dx(y^2/2) \ dx= int \ x/sqrt(x^21) \ dx = int \ d/dx( sqrt(x^21)) \ dx implies y^2= sqrt(x^21) C y^2= C sqrt(x^21) y= pm sqrt ( C sqrt(x^21) ) applying the IV y(0) = 1 1= pm sqrt ( C sqrt(1) ) so we can disregard the ve sign 1= sqrt ( C 1 ) implies C = 2
(x 2 x 1) dy (y 2 y 1) dx = 0 (x 2 x Consider the following statements 1) The function f(x) = sin x decre What is the number of arbitrary constants in the particular solution oSince the integral of 1x^2 is arctanX, I get arctanX C = arctanY C And I don't know what to do from here Please help!X^2 2 y^2 = 1 Natural Language;
I realize you must divide the dy/dx coefficient through everything but how did it simplify to that term?Dy/dx Py = Q where P = (1/x) and Q =3/x To solve it , multiplying the whole equation by I F = exp (Int (1/x)) = x, so that itX 1y y 1x = 0x 1y = −y 1x Square on both sides(x 1y )2 = (−y 1x )2x2(1y)= (−y)2(1x)x2(1y)= y2(1x)x2 x2y = y2 y2xx2 −y2 = y2x−x2y(xy)(x−y) = −xy(x−y)xy = −xyx = −xy−yx = −y(x1)y = −x1x Differentiate y with respect to x using Chain rule, we get ;dxdy = (x1)2−1(x1)−(−x)1 dxdy = −(x1)21
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Given dy/dx=x/(1x^2) Multiply both sides by dx dy=x/(1x^2)dx Integrate intdy=intx/(1x^2)dx intdy=1/2int(2x)/(1x^2)dx intdy = 1/2(du)/u y = 1/2ln(u) C y = 1/2ln(x^21)C y = ln(sqrt(x^21))C CalculusMath Advanced Math Advanced Math questions and answers Solve the initial value problems x^2 dy/dx = 4x^2 x 2/ (x 1) (y 1), y (1) = 1 x^2 dy/dx 3xy = x^4 ln (x) 1, y (1) = 0 Solve the Equation dy/dx y/x 2 = 5 (x 2)y^1/2 Determine whether the equation is exact if it is then solve it 2/Squareroot 1 x^2 y cos (xy) dx x cos (xy) y^1/3 dy = 0 Solve the Equation dy/dx = x sec (y/x) y/x solve the differential equation (x^21) dy/dx2 (x2) y=2 (x1) Share with your friends Share 5
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Find dy/dx y=1/(x^2) Differentiate both sides of the equation The derivative of with respect to is Differentiate the right side of the equation Tap for more steps Apply basic rules of exponents Tap for more steps Rewrite as Multiply the exponents in Tap for more stepsAnswer to Solve the initial value problem dy/dx = (y^2 1)/(x^2 1), y(2) = 2 By signing up, you'll get thousands of stepbystep solutions to1 Inform you about time table of exam 2 Inform you about new question papers 3 New video tutorials information
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Please help Thanks in advance We have x2=2xy 3y2 = 0 Are there supposed to be 2 equal signs in this expression or is it x2 calc xy^2 2xy = 8, then, at the point (1,2) y' is? Solve the differential equation dy = cos x (2y cosec x) dx given that y = 2 when x = π/2 asked in Class XII Maths by nikita74 ( 1,017 points) differential equationsHomework Equations The Attempt at a Solution
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$$\frac{du}{u^{2}{\left(x \right)} 2 u{\left(x \right)} 1} = \frac{dx}{2 x}$$ Take the integrals from the both equation sides the integral of the left side by u, the integral of the right side by x $$\int \frac{1}{u^{2} 2 u 1}\, du = \int \frac{1}{2 x}\, dx$$ Detailed solution of the integral with u Detailed solution of theTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW `(1x^2) dy/dxxy=1`Weekly Subscription $199 USD per week until cancelled Monthly Subscription $699 USD per month until cancelled Annual Subscription $2999 USD per year until cancelled
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Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, musicAsked in Class XII Maths by rahul152 (2,8 points) Solve the differential equation dy/dx = 1xy 2 xy 2, when y = 0, x = 0This problem has been solved!
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Answer to Find \\frac{dy}{dx} y = (1 x^2)^{\\sqrt x} (Note Express the answer in terms of x only) By signing up, you'll get thousands ofDivide both sides by x(1 — y) to put all factors with x on the left and all factors with y on the right (1 — x^2)/x dx = y(1 y)/ (1 — y) dy Now integrate both sides You'll need to use polynomial division to rewrite both sides as a polynomial plus a proper rational expression first answered by Prerna01 (521k points) selected by RahulYadav Best answer Given as x2/a2 y2/b2 = 1 Differentiating the equation on both sides with respect to x, 2x/a2 (2y/b2) (dy/dx) = 0 dy/dx =
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A first order Differential Equation is Homogeneous when it can be in this form dy dx = F ( y x ) We can solve it using Separation of Variables but first we create a new variable v = y x v = y x which is also y = vx And dy dx = d (vx) dx = v dx dx x dv dx (by the Product Rule) Which can be simplified to dy dx = v x dv dxSum Rule \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx I came across a question solve $$\frac{dy}{dx} = \frac{1}{xy^2},\qquad y(2) = 0$$ I have no clue how to proceed I have suspicion that this equation has no closed form solution and may be there was a mistake in the question Course is a basic course on ODEs I will appreciate solution/hints/further reading
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On substituting the value of x = 1 and y = 1, we get ⇒ dy / dx = not defined anything divided by 0 is not defined Therefore, the differentiation of the expression 3x 2 2xy y 2 = 2 at x = 1 is not defined Explore math program Explore coding program Pick your preferred day & timeSee the answer (x^2 1)dy/dx 2y = (x1)^2 put in standard form, dy/dx 2/ (x^2 1) *y = x1 /x1 how did that happen??Simple and best practice solution for (2xy)dy(x^2y^21)dx=0 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework If it's not what You are looking for type in the equation solver your own equation and let us solve it
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Let's simplify it First dy/dx = (y/x 1)/(y/x 1) Taking y = vx dy/dx = v xdv/dx Therefore, dx/x = (v 1)dv / (v^2 1) Integrating we get log (1/x) logc = arctan (y/x) 1/2 logMultiply − 2 2 by − 1 1 Since − 1 1 is constant with respect to x x, the derivative of − 1 1 with respect to x x is 0 0 Simplify the expression Tap for more steps Multiply 1 x 2 1 x 2 by 0 0 Add 2 x − 3 2 x 3 and 0 0 Rewrite the expression using the negative exponent rule b−n = 1 bn b n = 1Simple one The given differential equation can be written as (dy/dx) (y/x) = 3/x which in now in standard form of first order linear differential equation ;
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Help is appreciated Edit(1 X) (1 Y2) Dx (1 Y) (1 X2) Dy = 0 CBSE CBSE (Science) Class 12 Question Papers 1851 Textbook Solutions Important Solutions 4562 Question Bank Solutions Concept Notes & Videos 725 Time Tables 18 Syllabus Advertisement Remove all adsIf y = 2^x, find dy/dx Q If y = 2^x, find dy/dx ANSWER 1) Take Logs of both sides of our equation y = 2^x So we get log (y)=log (2^x) 2) Apply relevant log rule to rhs Log rule log (a^b) = b log (a) nb the dot between b and log (a) represents x / multiply / times ) So we get log (y) = x log (2)
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Instead follow this method, Substitute y^2 = u and solve, ie keeping u as a variable find du/dx in terms of dy/dx and substitute also substitute the value of y in the expression sandeep 15 Points 2 years ago Ans dy/dx = (x^2y^21)/2xy y^2 = v rewrite the equation in the form of linear equation Then solve it (1x^2)dy/dx xy = 1/ (1x^2) the ans given is y= x/ (1x^2) C / ( sqrt rt (1x^2) ) , my ans is different , which part is wrong ?Y 1 y 1 dy = x 2 x 3 dx 3 To evaluate the integrals Z y 1 y 1 dy = Z x 2 x 3 dx we need usubstitution on both sides On the LHS, let u = y 1 and then du = dy and y = u1 On the RHS we need another variable name, so let w = x 3 and then dw = dx and x = w 3 Substituting (01
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Homework Solutions MATH 32B2 (18W) Problem 10 () Sketch the region Dbetween y= x2 and y= x(1 x) Express Das a simple region and calculate the integral of f(x;y) = 2yover DSolutions to x^2 dy^2 = 1 and x^2 dy^2 = 1 Recall from the Pell's Equation page that Pell's equation is $x^2 dy^2 = N$ where $d, N \in \mathbb{Z}$ We noted that Pell's equation has at most finitely many solutions if $d < 0$ or $d$ is a perfect square We will now look at a special type of Pell's equation and when it has solutions, namely (1) Solve the differential equation (x^2 1) dy/dx 2xy = 2/(x^2 1), where x ∈ ( ∞, 1) ⋃ (1, ∞) askedMar 13in Differential Equationsby Yaad(352kpoints) differential equations class12 0votes 1answer Solve the following differential equation 2x^2 dy/dx 2xy y^2 = 0
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