Mth401 Assignment No.1 Solution Fall 2024 Section: Zulfiqar Ahmad Noor

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MTH401 - Differential Equations Assignment No.1 Solution Section In charge: Zulfiqar Ahmad Noor Question # 1 Consider the following differential equation: (2𝑥𝑦 + 3)𝑑𝑥 + (𝑥2 + 4𝑦)𝑦 = 0 (a) Verify whether this differential equation is exact. (Show steps) (b) If it is exact, find the general solution of the given differential equation _________________________________________________________ Solution: (a) Verify whether this differential equation is exact. (Show steps) A differential equation of the form M (x , y) dx + N (x , y) dy = 0 is exact if ∂M/∂y=∂N/∂x For our equation: M(x,y) = 2xy + 3 N(x,y) = x2 + 4y Find ∂M/∂y: ∂M/∂y = ∂/∂y (2xy + 3) = 2x Find ∂N/∂x: ∂N/∂x=∂/∂x(x2+4y) = 2x Since ∂M/∂y=∂N/∂x = 2x, the differential equation is exact. (b) The general solution: To find the general solution we need to find S and T then solution is S+T=C S = ∫ Mdx = ∫ (2xy + 3)dx S = ∫ Mdx = ∫ (2xy) dx + 3 ∫ dx S = 2y x^2/2 + 3x S = x^2 y + 3x T = Integrate that portion of N w.r.t y that is free from x T = ∫ (4y) dy T = 4 ∫y dy T = 4 y^2/2 = 2y2 the general solution of the differential equation is: S +T = C x2y + 3x + 2y2 = C Where C is a constant. Question # 2: For each of the differential equations in the given table: a) Identify the type of differential equation and state the method you would use to solve it. b) Solve any one differential equation of your choice from the table. Sr # Differential Equations Suggested Method 1 dy/dx = xy2 2 (3x2+y)dx + (x+ 4y3)dy =0 3 y’ - 2/3y = x2 4 dy/dx + y = ex 5 dy/dx + y = y3x 6 (2x + 3y) dx + (3x + 4y)dy = 0 7 dy/dx - y/x = x3 8 y’+ y = cos x 9 dy/dx + 3y = y2 cos x _________________________________________________________ Solution: Sr # Differential Equations type of differential equation Suggested Method 1 dy/dx = xy2 Separable Separate variables and integrate both sides 2 (3x2+y)dx + (x+ 4y3)dy =0 Potentially Exact Check for exactness if exact, solve by finding a potential function; if not exact, find an integrating factor 3 y’ - 2/3y = x2 First-order linear Use integrating factor e∫-2/3 dx 4 dy/dx + y = ex First-order linear Use integrating factor e∫xdx 5 dy/dx + y = y3x Bernoulli equation Convert by setting v = y-2 and solve as a linear equation 6 (2x + 3y) dx + (3x + 4y)dy = 0 Potentially Exact Check for exactness if exact, solve by finding a potential function; if not exact, find an integrating factor 7 dy/dx - y/x = x3 First-order linear Use integrating factor e∫-1/x dx=1/x 8 y’+ y = cos x First-order linear Use integrating factor e∫xdx=ex 9 dy/dx + 3y = y2 cos x Bernoulli equation Convert by setting v = y-1 and solve as a linear equation Solution of self-chosen equation: Let's solve equation #4: dy/dx + y = ex This is a first-order linear differential equation. The standard form of a linear first-order differential equation is: dy/dx + P(x)y = Q(x) In our case: P(x) =1 Q(x) = ex Step 1: Find the integrating factor For a linear equation dy/dx + P(x)y = Q(x) the integrating factor is: μ(x)=e∫P(x)dx Here, P(x) = 1, so we have: μ(x) = e∫1dx = ex Step 2: Multiply the entire differential equation by the integrating factor Multiply both sides of the equation dy/dx + y = ex by ex: ex. dy/dx + ex .y = ex . ex Simplifying: ex dy/dx + exy = e2x Step 3: Observe the left-hand side The left-hand side of the equation is the derivative of y⋅μ(x)y \cdot \mu(x)y⋅μ(x), where μ(x)=ex\mu(x) = e^xμ(x)=ex. This can be written as: d/dx (exy) = e2x Step 4: Integrate both sides Now, integrate both sides with respect to x: ∫ d/dx (ex.y)dx = ∫e2x dx On the left-hand side, the integral is straightforward: exy = 1/2.e2x + C Where C is the constant of integration. Step 5: Solve for y Now, solve for y by dividing both sides by ex: y = e2x/2ex + C/ex Simplify the expression: y = ex/2 + Ce-x Final Solution: The general solution to the differential equation is: y = ex/2 + Ce-x Where C is the constant of integration. The end

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AOA sir G mth403 Ki assignment ka solution bhi upload krdei plz section in-charge arooba Fatima plz sir G