y’+2y = c*(-2)*exp(-2x) –exp(-x) + 2c*exp(-2x) +2exp(-x) =

= exp(-x); yes, it’s true!

y(1) =c*exp(-2) +exp(-1) =7, hence

c= (7-exp(-1))*exp(2) = 7*exp(2) –exp(1);

2\♣ y’ = k*exp(kx); y’’ = k^2*exp(kx);

y’’ -17y’ +72y = k^2*exp(kx) –17*k*exp(kx) +72*exp(kx) =0;

k^2 –17k +72 =0; k=(17 ±√(17^2 –4*72))/2, hence k1=8, k2=9;

3/♦ thus u’ =exp(3.3t) /exp(1.6u);

exp(1.6u)*du = exp(3.3t)*dt, hence

exp(1.6u)/1.6 = exp(3.3t)/3.3 +C;

exp(1.6*1.4)/1.6 = exp(3.3*0)/3.3 +C;

C= exp(2.24)/1.6 –1/3.3;

Therefore ln{exp(1.6u)/1.6} = ln{exp(3.3t)/3.3 +C};

1.6*u –ln(1.6) = ln{exp(3.3t)/3.3 +C}, hence

u(t) = ln{(exp(3.3t)/3.3 +C)/1.6}/1.6, where

C= exp(2.24)/1.6 –1/3.3; simplify!

4\■ thus 4*(2yy’)=x, thence 4(y^2)’ =(x^2 /2)’, hence

4y^2=x^2/2 +4C; y=√(x^2/8 +C);

y(8) = 10 =√(8^2/8 +C), hence C=100-8 = 92;

y(x) =√(x^2/8 +92);

5/▲thus 7xdx = 4y(x^(2)+1)^(1/2) dy;

7xdx/√(x^2+1) = 4ydy;

7√(x^2+1) +2C = 2y^2; y=√(3.5 √(x^2+1) +C);

y(0) =6 =√(3.5 √(0^2+1) +C), hence C=36-3.5 =32.5;

y=√(3.5 √(x^2+1) +32.5);

♥ I love that lovely hairy wart on your neck;