Let a curve y=f(x) pass through the points (0,5) and (loge2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=0, then k is equal to
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Step-by-Step Solution
Step 1: Separate the variables
The given differential equation is 2(3+y)e2xdx−(7+e2x)dy=0. To solve this, we need to separate the variables x and y. We rearrange the equation to group all terms involving y with dy and all terms involving x with dx. This gives us a separable differential equation.
Step 2: Integrate both sides
Now that the variables are separated, we integrate both sides of the equation. The integral of 3+y1 with respect to y is log∣3+y∣. For the right side, we can use a substitution u=7+e2x, which means du=2e2xdx. So the integral becomes ∫u1du=log∣u∣.
Step 3: Solve the integrals
After integrating both sides, we get log∣3+y∣=log∣7+e2x∣+C, where C is the constant of integration. We can rewrite the constant C as logA for some constant A>0.
Step 4: Simplify the general solution
Using the logarithm property loga+logb=log(ab), we can combine the terms on the right side: log∣7+e2x∣+logA=log∣A(7+e2x)∣. Then, by exponentiating both sides, we remove the logarithms, yielding 3+y=A(7+e2x).
Step 5: Use the initial condition to find A
The curve passes through the point (0,5). We substitute x=0 and y=5 into the general solution to find the value of the constant A. This gives 3+5=A(7+e0), which simplifies to 8=A(7+1), so 8=8A, meaning A=1.
Step 6: Find k using the second point
Now that we have the specific solution 3+y=7+e2x, we use the second point (loge2,k). Substitute x=loge2 and y=k into the equation. We use the logarithm property elogeM=M and alogb=logba. So e2loge2=eloge22=eloge4=4. This leads to 3+k=7+4, which simplifies to 3+k=11, and thus k=8.