2.1875   ODE No. 1875

1. Problem in Latex
2. Mathematica input
3. Maple input

$$\{f(t)(ax(t)+by(t))+x^{\prime }(t)=g(t),f(t)(cx(t)+dy(t))+y^{\prime }(t)=h(t)\}$$ ✗ Mathematica : cpu = 0.0055809 (sec), leaf count = 0 , could not solve

DSolve[{f[t]*(a*x[t] + b*y[t]) + Derivative[1][x][t] == g[t], f[t]*(c*x[t] + d*y[t]) + Derivative[1][y][t] == h[t]}, {x[t], y[t]}, t]

✓ Maple : cpu = 1.095 (sec), leaf count = 2606

$$\left\{\{x\left(t\right)={\mathrm{e}}^{\int -\frac{f\left(t\right)}{2a+2d}(\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-a^4-4a^{2}bc+2a^2{d}^2-8abcd-4bc{d}^2-d^4}\right)\sqrt{-a^4-4a^{2}bc+2a^2{d}^2-8abcd-4bc{d}^2-d^4}+{(a+d)}^2)\mathrm{d}t}\mathit{\_}\mathit{C}\mathit{2}+{\mathrm{e}}^{\int -\frac{f\left(t\right)}{2a+2d}(\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-a^4-4a^{2}bc+2a^2{d}^2-8abcd-4bc{d}^2-d^4}\right)\sqrt{-a^4-4a^{2}bc+2a^2{d}^2-8abcd-4bc{d}^2-d^4}+{(a+d)}^2)\mathrm{d}t}\mathit{\_}\mathit{C}\mathit{1}-2\frac{a+d}{\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}(\int -\frac{dg\left(t\right){\left(f\left(t\right)\right)}^2-bh\left(t\right){\left(f\left(t\right)\right)}^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}g\left(t\right)\right)f\left(t\right)-g\left(t\right)\frac{\mathrm{d}}{\mathrm{d}t}f\left(t\right)}{{\left(f\left(t\right)\right)}^2}{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{2a+2d}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}{(-\tan \left(\frac{\int f\left(t\right)\mathrm{d}t\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{2a+2d}\right)+\tan \left(1/2\frac{(-1+(a+d)\int f\left(t\right)\mathrm{d}t)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{{(a+d)}^2}\right))}^{-1}\mathrm{d}t{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(1/2\frac{(-1+(a+d)\int f\left(t\right)\mathrm{d}t)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{{(a+d)}^2}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}-\int -\frac{dg\left(t\right){\left(f\left(t\right)\right)}^2-bh\left(t\right){\left(f\left(t\right)\right)}^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}g\left(t\right)\right)f\left(t\right)-g\left(t\right)\frac{\mathrm{d}}{\mathrm{d}t}f\left(t\right)}{{\left(f\left(t\right)\right)}^2}{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(1/2\frac{(-1+(a+d)\int f\left(t\right)\mathrm{d}t)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{{(a+d)}^2}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}{(-\tan \left(\frac{\int f\left(t\right)\mathrm{d}t\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{2a+2d}\right)+\tan \left(1/2\frac{(-1+(a+d)\int f\left(t\right)\mathrm{d}t)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{{(a+d)}^2}\right))}^{-1}\mathrm{d}t{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{2a+2d}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}){\mathrm{e}}^{\frac{1}{2a+2d}(-\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(1/2\frac{(-1+(a+d)\int f\left(t\right)\mathrm{d}t)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{{(a+d)}^2}\right)\mathrm{d}t-\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}{2a+2d}\right)\mathrm{d}t-2\int f\left(t\right)\mathrm{d}t{(a+d)}^2)},y\left(t\right)=-\frac{1}{b(a+d)f\left(t\right)}({\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}{\mathrm{e}}^{\frac{1}{2a+2d}(-\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t-\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t-2\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}f\left(t\right)(a+d)(a^2-\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}-d^2)\int -\frac{dg\left(t\right){\left(f\left(t\right)\right)}^2-bh\left(t\right){\left(f\left(t\right)\right)}^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}g\left(t\right)\right)f\left(t\right)-g\left(t\right)\frac{\mathrm{d}}{\mathrm{d}t}f\left(t\right)}{{\left(f\left(t\right)\right)}^2}{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}{(-\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)+\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right))}^{-1}\mathrm{d}t-{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}{\mathrm{e}}^{\frac{1}{2a+2d}(-\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t-\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t-2\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}f\left(t\right)(a+d)(a^2-\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}-d^2)\int -\frac{dg\left(t\right){\left(f\left(t\right)\right)}^2-bh\left(t\right){\left(f\left(t\right)\right)}^2+\left(\frac{\mathrm{d}}{\mathrm{d}t}g\left(t\right)\right)f\left(t\right)-g\left(t\right)\frac{\mathrm{d}}{\mathrm{d}t}f\left(t\right)}{{\left(f\left(t\right)\right)}^2}{\mathrm{e}}^{\frac{1}{2a+2d}(\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\int f\left(t\right)\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\mathrm{d}t+\int f\left(t\right)\mathrm{d}t{(a+d)}^2)}{(-\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)+\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right))}^{-1}\mathrm{d}t+\frac{1}{2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}(\mathit{\_}\mathit{C}\mathit{2}f\left(t\right)(a^2-\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}-d^2){\mathrm{e}}^{\int -\frac{f\left(t\right)}{2a+2d}(\tan \left(\frac{-1+(a+d)\int f\left(t\right)\mathrm{d}t}{2{(a+d)}^2}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}+{(a+d)}^2)\mathrm{d}t}+\mathit{\_}\mathit{C}\mathit{1}f\left(t\right)(a^2-\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}-d^2){\mathrm{e}}^{\int -\frac{f\left(t\right)}{2a+2d}(\tan \left(\frac{\int f\left(t\right)\mathrm{d}t}{2a+2d}\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}\right)\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}+{(a+d)}^2)\mathrm{d}t}-2g\left(t\right)(a+d)))\frac{1}{\sqrt{-{(a+d)}^2(a^2-2ad+4bc+d^2)}}\}\right\}$$