7.15
[tex]\dfrac y{x^2}\,\mathrm dx-\dfrac{xy+1}x\,\mathrm dy=0[/tex]
This is a Bernoulli equation in disguise. Solve for [tex]\frac{\mathrm dx}{\mathrm dy}=x'[/tex]:
[tex]x'=\dfrac{x(xy+1)}y=x^2+\dfrac xy[/tex]
Divide through both sides by [tex]x^2[/tex]:
[tex]x^{-2}x'=1+\dfrac1{xy}[/tex]
Substitute [tex]v=x^{-1}[/tex] and [tex]v'=-x^{-2}x'[/tex], and the resulting ODE is linear in [tex]v[/tex].
[tex]-v'=1+\dfrac vy\implies v'+\dfrac vy=-1[/tex]
Multiply both sides by [tex]y[/tex], integrate both sides, solve for [tex]v[/tex], then for [tex]x[/tex]:
[tex]yv'+v=(yv)'=-y\implies yv=-\dfrac{y^2}2+C\implies \boxed{x(y)=\dfrac{2y}{C-y^2}}[/tex]
8.15
[tex]yy'=-\dfrac x2[/tex]
This is separable; write [tex]y'=\frac{\mathrm dy}{\mathrm dx}[/tex], then
[tex]y\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac x2\implies y\,\mathrm dy=-\dfrac x2\,\mathrm dx[/tex]
Integrate both sides to get
[tex]\dfrac{y^2}2=-x^2+C[/tex]
Given that [tex]y=2[/tex] when [tex]x=4[/tex], we find
[tex]\dfrac{2^2}2=-4^2+C\implies C=18[/tex]
so the particular solution is
[tex]\boxed{\dfrac{y(x)^2}2=18-x^2}[/tex]
10.15
The image is a bit blurry, but it looks like the equation is
[tex]xy'''-y''+\dfrac1x=0[/tex]
Substitute [tex]v=y''[/tex], so that [tex]v'=y'''[/tex] and the equation is linear in [tex]v[/tex]:
[tex]xv'-v=-\dfrac1x[/tex]
Divide through both sides by [tex]x^2[/tex]:
[tex]\dfrac{v'}x-\dfrac v{x^2}=\left(\dfrac vx\right)'=-\dfrac1{x^3}[/tex]
Integrate both sides and solve for [tex]v[/tex]:
[tex]\dfrac vx=\dfrac1{2x^2}+C[/tex]
[tex]v=\dfrac1{2x}+Cx[/tex]
Solve for [tex]y[/tex] by integrating both sides twice:
[tex]y''=\dfrac1{2x}+Cx[/tex]
[tex]y'=\dfrac{\ln x}2+C_1x^2+C_2[/tex]
[tex]\boxed{y(x)=\dfrac{x(\ln x-1)}2+C_1x^3+C_2x+C_3}[/tex]
11.15
[tex]y''y^3+25=0[/tex]
Unfortunately, I'm not sure how to tackle this one just yet...
12.15
[tex]y''-y'=3x^2-2x+1[/tex]
You could use undetermined coefficients here, but since the equation is free of [tex]y[/tex], you can substitute [tex]v=y'[/tex] and [tex]v'=y''[/tex] to reduce the order and get a linear equation.
[tex]v'-v=3x^2-2x+1[/tex]
Multiply through both sides by [tex]e^{-x}[/tex], integrate, solve ... you know the drill:
[tex]e^{-x}v'-e^{-x}v=(e^{-x}v)'=(3x^2-2x+1)e^{-x}[/tex]
[tex]e^{-x}v=-(3x^2+4x+5)e^{-x}+C[/tex]
[tex]v=-(3x^2+4x+5)+Ce^x[/tex]
[tex]\boxed{y(x)=-(x^3+2x^2+5x)+C_1e^x+C_2}[/tex]
13.15
[tex]y'''+4y''+4y'=(9x+15)e^x[/tex]
We could reduce the order again, but that won't avoid having to solve a higher-order ODE like in the previous examples. The homogeneous equation
[tex]y'''+4y''+4y'=0[/tex]
has characteristic equation
[tex]r^3+4r^2+4r=r(r^2+4r+4)=r(r+2)^2=0[/tex]
with roots [tex]r=0[/tex] and [tex]r=-2[/tex] (with multiplicity 2), and thus the characteristic solution is
[tex]y_c=C_1+C_2e^{-2x}+C_3xe^{-2x}[/tex]
For the particular solution, assume an ansatz
[tex]y_p=(a_0+a_1x+a_2x^2)e^x[/tex]
with derivatives
[tex]{y_p}'=(a_0+a_1+(a_1+2a_2)x+a_2x^2)e^x[/tex]
[tex]{y_p}''=(a_0+2a_1+2a_2+(a_1+4a_2)x+a_2x^2)e^x[/tex]
[tex]{y_p}'''=(a_0+3a_1+6a_2+(a_1+6a_2)x+a_2x^2)e^x[/tex]
Substitute these into the ODE and simplify to end up with
[tex](9a_0+15a_1+14a_2+(9a_1+30a_2)x+9a_2x^2)e^x=(9x+15)e^x[/tex]
Matching up coefficients tells us
[tex]9a_2=0\implies a_2=0[/tex]
[tex]9a_1+30a_2=9\implies a_1=1[/tex]
[tex]9a_0+15a_1+14a_2=15\implies a_0=0[/tex]
So the particular solution is
[tex]y_p=xe^x[/tex]
and the general solution to the ODE is
[tex]\boxed{y(x)=C_1+C_2e^{-2x}+C_3xe^{-2x}+xe^x}[/tex]
14.15
[tex]y''+y=2\cos5x+3\sin5x[/tex]
The characteristic equation is
[tex]r^2+1=0[/tex]
with roots [tex]r=\pm i[/tex], which admits the characteristic solution
[tex]y_c=C_1\cos x+C_2\sin x[/tex]
Assume the ansatz
[tex]y_p=a\cos5x+b\sin5x[/tex]
with second derivative
[tex]{y_p}''=-25a\cos5x-25b\sin5x[/tex]
Substitute these into the ODE, simplify, and solve for [tex]a,b[/tex]:
[tex]-24a\cos5x-24b\sin5x=2\cos5x+3\sin5x[/tex]
[tex]-24a=2\implies a=-\dfrac1{12}[/tex]
[tex]-24b=3\implies b=-\dfrac18[/tex]
Then the general solution is
[tex]\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac{\cos5x}{12}-\dfrac{b\sin5x}8}[/tex]
