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3

For this specific case the exact solution can be found. Clear["Global`*"] eqns = {y'[t] == a y[t], y == 1}; sol = DSolve[eqns, y, t][] (* {y -> Function[{t}, E^(a t)]} *) Verifying the solution, eqns /. sol (* {True, True} *) a[x_, const_] = a /. Solve[(y[x] /. sol) == const, a][] /. C -> 0 (* Log[const]/x *) With[{const = 1.15}, ...

2

My method is uglier than just extracting a spline like with C. E's answer. It's possible find points on this contour with an NMinimize and use an Interpolation to get the curve as a function of $x$ (see func below): sol = ParametricNDSolve[{y'[t] == a y[t], y == 1}, y, {t, 0, 10}, {a}]; fn = y /. sol; (* Set the target contour *) target = 1.15; (* ...

2

What you may not know is that the notebook interface is a bit like a web browser. Whatever complicated interface the web browser is showing, you can always just right-click and show the HTML source code for it. It's not delivered as a bunch of pixels, and similarly, graphics (to be distinguished from actual images, including rasterized graphics) are just ...

1

You can always do this if you are not tied to a series: f = InverseFunction[EllipticE[#1, -9.9] &] f[.656025] (*0.5*) or g = InverseFunction[Function[{x, y}, EllipticE[x, y]], 1, 2] g[.656025, -9.9] (*0.5*) This series works for smaller x Clear[k] f[x_, k_] = InverseSeries[Series[EllipticE[x, k], {x, 0, 20}]] // Simplify // Normal k = -9.9; f[....

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