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Calculus · real student question

Solve the differential equation dy/dx = ((2y + 3)/(4x + 5))^2 and give the general solution in implicit form.

Question

Solve

dydx=(2y+34x+5)2\frac{dy}{dx}=\left(\frac{2y+3}{4x+5}\right)^{2}

and give the general solution in implicit form.

Step-by-step solution

  1. Check that the equation is separable before choosing a method. Squaring the quotient keeps the yy-part and the xx-part in separate factors:

    dydx=(2y+3)2(4x+5)2,\frac{dy}{dx}=\frac{(2y+3)^{2}}{(4x+5)^{2}},

    which is exactly the form y′=g(y)h(x)y'=g(y)h(x). No integrating factor is needed — this is not linear in yy, so trying the linear-ODE machinery here would fail.

  2. Separate the variables. Divide by (2y+3)2(2y+3)^{2} and multiply by dxdx:

    dy(2y+3)2=dx(4x+5)2.\frac{dy}{(2y+3)^{2}}=\frac{dx}{(4x+5)^{2}}.

    Dividing by (2y+3)2(2y+3)^{2} costs you the constant solution y=−32y=-\tfrac32, which does satisfy the original equation (y′=0y'=0 and the right side is 00). It is a singular solution outside the general family, so note it now.

  3. Integrate the yy-side with the substitution u=2y+3u=2y+3. Then du=2 dydu=2\,dy, so dy=du2dy=\tfrac{du}{2} and

    ∫dy(2y+3)2=12∫u−2 du=−12u=−12(2y+3).\int\frac{dy}{(2y+3)^{2}}=\frac12\int u^{-2}\,du=-\frac{1}{2u}=-\frac{1}{2(2y+3)}.

    The factor 12\tfrac12 comes from the chain rule; forgetting it is the single most common error in linear-substitution integrals.

  4. Integrate the xx-side the same way with v=4x+5v=4x+5. Here dv=4 dxdv=4\,dx, so

    ∫dx(4x+5)2=14∫v−2 dv=−14v=−14(4x+5).\int\frac{dx}{(4x+5)^{2}}=\frac14\int v^{-2}\,dv=-\frac{1}{4v}=-\frac{1}{4(4x+5)}.

    The constant out front is 14\tfrac14 because the inner derivative is 44, matching the pattern ∫f(ax+b) dx=1aF(ax+b)\int f(ax+b)\,dx=\tfrac1a F(ax+b).

  5. Combine and absorb the constants. Equating the two antiderivatives and adding one arbitrary constant:

    −12(2y+3)=−14(4x+5)+C.-\frac{1}{2(2y+3)}=-\frac{1}{4(4x+5)}+C.

    Multiplying by −2-2 and renaming the constant gives the tidier implicit form

    12y+3=18x+10+C,\frac{1}{2y+3}=\frac{1}{8x+10}+C,

    since 24(4x+5)=12(4x+5)=18x+10\dfrac{2}{4(4x+5)}=\dfrac{1}{2(4x+5)}=\dfrac{1}{8x+10}.

  6. Verify by implicit differentiation. Differentiating 12y+3=18x+10+C\frac{1}{2y+3}=\frac{1}{8x+10}+C with respect to xx:

    −2y′(2y+3)2=−8(8x+10)2=−84(4x+5)2,-\frac{2y'}{(2y+3)^{2}}=-\frac{8}{(8x+10)^{2}}=-\frac{8}{4(4x+5)^{2}},

    so y′=(2y+3)2(4x+5)2y'=\frac{(2y+3)^{2}}{(4x+5)^{2}} ✓. A numerical spot check with C=0.3C=0.3 at x=1x=1 gives y′=0.09765625y'=0.09765625 from the implicit solution and 0.097656250.09765625 from the right-hand side.

Answer

12y+3=18x+10+Ctogether with the singular solution y=−32\frac{1}{2y+3}=\frac{1}{8x+10}+C\qquad\text{together with the singular solution }y=-\tfrac32

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