Q. QuestionWhat is the particular solution to the differential equation dxdy=x1+y with the initial condition y(e)=1?
Identify type of differential equation: Step 1: Identify the type of differential equation. We have dxdy=x1+y. This is a first-order linear differential equation.
Rearrange to separate variables: Step 2: Rearrange the equation to separate variables. 1+ydy=xdx. This step involves separating the variables y and x on different sides of the equation.
Integrate both sides: Step 3: Integrate both sides.Integrate left side: ∫1+ydy=ln∣1+y∣+C1.Integrate right side: ∫xdx=ln∣x∣+C2.
Combine constants and solve: Step 4: Combine the constants and solve for y.ln∣1+y∣=ln∣x∣+C, where C=C2−C1.Remove the absolute values (assuming x and 1+y are positive): 1+y=e(ln∣x∣+C)=x⋅eC.
Apply initial condition: Step 5: Apply the initial condition to find C. Given y(e)=1, plug in x=e: 1+1=e⋅eC. 2=e(1+C). Taking natural log on both sides: ln(2)=1+C. C=ln(2)−1.
Write particular solution: Step 6: Write the particular solution using the value of C.y=x⋅e(ln(2)−1)−1.Simplify: y=x⋅(e2)−1.
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