Maths 1:Ans=1/10x^2+c 2:Ans=sinx e^x?

1.
3x² dx - 15x dy = 0
15x dy = 3x² dx
dy = (3x² dx) / (15x)
dy = (1/5)x dx
∫dy = ∫(1/5)x dx
y = (1/5) (1/2)x² + C
y = (1/10)x² + C


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2.
Method 1 :

Integrated by parts : ∫u dv = u v - ∫v du

∫(sinx + cosx) e^x dx
= ∫e^x [(sinx + cosx) dx]
= ∫e^x d(-cosx + sinx)
= e^x (-cosx + sinx) - ∫(-cosx + sinx) d(e^x)
= e^x (-cosx + sinx) - ∫(-cosx + sinx) e^x dx
= e^x (-cosx + sinx) - ∫e^x [(-cosx + sinx) dx]
= e^x (-cosx + sinx) - ∫e^x d(-sinx - cosx)
= e^x (-cosx + sinx) + ∫e^x d(sinx + cosx)
= e^x (-cosx + sinx) + e^x (sinx + cosx) - ∫(sinx + cosx) d(e^x)
= e^x (-cosx + sinx + sinx + cosx) - ∫(sinx + cosx) e^x dx
= 2 e^x sinx - ∫(sinx + cosx) e^x dx

Rearrange, we get :
2 ∫(sinx + cosx) e^x dx = 2 e^x sinx + C₁
Hence, ∫(sinx + cosx) e^x dx = e^x sinx + C


Method 2 :
(d/dx)(e^x sinx)
= e^x [d(sinx)/dx] + sinx [d(e^x)/dx]
= e^x cosx + e^x sinx
= (sinx + cosx) e^x

Hence, (sinx + cosx) e^x dx = d(e^x sinx)

Take integration for the both sides:
∫(sinx + cosx) e^x dx = ∫d(e^x sinx)
∫(sinx + cosx) e^x dx = e^x sinx + C

Well,

1:
If 3x^2 dx -15x dy=0
then
3x^2 dx = 15x dy
dy/dx = 3x^2 /15x
dy/dx = x/5
therefore
y = x^2 /10 + C

2: we see that :
( sinx e^x ) ' = sinx e^x + cosx e^x
= (sinx + cosx) e^x
therefore
∫ (sinx + cosx) e^x = sinx e^x

Hint :
the integral
∫ (sinx + cosx) e^x
is under the form
∫ u dv + ∫ v du ==> the integral is ( uv )

hope it' ll help !!

1) this DE is separable
3x² dx -15xdy=0
3x² dx = 15x dy
3x²/15x dx = dy
x/5 dx = dy
∫ x/5 dx = ∫ dy
y = 1/10 x² + C

2) circular integration by parts twice
integrate by parts
A = ∫ (sin(x)+cos(x))e^x dx
du = e^x dx, v = sin(x)+cos(x)
u = e^x, dv = (cos(x)-sin(x)) dx
A = ∫ v du = uv - ∫ u dv
= e^x(sin(x)+cos(x))
- ∫ e^x(cos(x)-sin(x)) dx
= e^x(sin(x)+cos(x)
+ ∫ e^x(sin(x)-cos(x)) dx

2nd term by parts again
du = e^x dx, v = sin(x)-cos(x) dx
u = e^x, dv = cos(x)+sin(x) dx
∫ v du = uv - ∫ u dv
= e^x(sin(x)-cos(x))
- ∫ e^x(sin(x)+cos(x))dx

A = e^x(sin(x)+cos(x))
+ ∫ e^x(sin(x)-cos(x)) dx
= e^x(sin(x)+cos(x))
+ e^x(sin(x)-cos(x))
- A

2A = 2e^x sin(x)
A = e^x sin(x) + C

A' = e^x sin(x) + e^x cos(x)
= e^x(sin(x)+cos(x))
QED

Michael shows that instead of integrating you can show that the differential of the answer is the desired integrand. Or he modifies integration by parts to simplify the problem. Adam shows Michael's first technique.
Both avoid the mess for #2.

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